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PHYSICS SEMESTER ONE UNIT 7 UNIT 7: ROTATION AND EQUILIBRIUM Relationship to Labs: This unit is supported by Lab 6: Torque and Rotational Equilibrium (kit-based lab). So far, we have covered motion of objects moving in a straight line, changing direction or going around a curve. Now, imagine trying to apply the linear motion analysis to a spinning object. All of the parts of the spinning object are moving in different directions, but the object itself may be stationary. We need a new analysis system to cover items that are spinning about an axis. Rotation is the motion of an object about a fixed axis. Rotation covers everything from spinning wheels and planets, to the use of a wrench on a bolt. When we get to the equilibrium portion of this section, we will see that we can use the linear and rotation properties to solve complicated problems where nothing is rotating. Rotation has several terms that have similar properties to terms in linear motion. Distance gets replaced with angle, speed with angular speed, acceleration with angular acceleration, mass with moment of inertia, force with torque, kinetic energy with rotational kinetic energy, and momentum with angular momentum. The relationships between the rotational terms are identical to the relationships between the linear motion terms. Furthermore, we can often convert linear motion expressions to rotational motion expressions; though expressions in one system are often simpler than the other. The axis of rotation is the point about which all other points rotate. A point, distance r from an axis of rotation, will move through an arc of length s. The arc length for the motion of the point rotating through s the angle θ (in radians) is s r Imagine an object that moves in a complete circle. The object rotates through the angle of 2 θ r The arc length around the circle is the circumference sC 2 r r This works for all angles. We can rearrange this to get Creative Commons Attribution 3.0 Unported License 1 PHYSICS SEMESTER ONE UNIT 7 s r This shows that the angle is the distance (arc length) divided by the radius. The angle (in radians) is considered dimensionless (unit-free) because it is a ratio of two lengths. We will now define the angular speed ω as ave , average angular speed t or d , the instantaneous angular speed dt Using our definition for the arc length (distance), the instantaneous speed is v ds dt dr dt r d dt r v r Creative Commons Attribution 3.0 Unported License = 4.0 s-1 object is 2.0 m/s. Find the angular speed of the point moving around the axis. = (2.0 m/s)/(0.50 m) The distance from a point to its axis of rotation is 0.50 m and the speed of the = v/r Quick Question equation: the units s-1, or radians per second. r = 0.50 m, v = 2.0 m/s quotient formed by the speed divided by the radius. The angular speed has terms: As in the relation between angle and distance, the angular speed is the The point moves through 4.0 radians per second. or 2 PHYSICS SEMESTER ONE UNIT 7 Units Rotation rates are often expressed in cycles per second (hertz, Hz) or revolutions per minute (rpm). A cycle is equivalent to a revolution. In one complete revolution, an object rotates through 2π radians. Similarly, in rotating through 1 radian an object moves through 1/2π. To convert from radians to revolutions or cycles: multiply by 1/2π To convert from revolutions or cycles to radians: multiply by 2π To convert from 7.00 rpm to radians per seconds 7.00 rpm 7.00 revolutions 2 radians 1 min min revolution 60 s 0.733 s 1 0.733 radians per second To convert from 2.0106 Hz to radians per seconds 2.0 106 Hz 2.0 106 2.0 106 cycles s cycles 2 radians s 1 cycle 1.3 107 s 1 1.3 107 radians per second Creative Commons Attribution 3.0 Unported License 3 PHYSICS SEMESTER ONE UNIT 7 MOMENT OF INERTIA Moment of inertia in rotating systems is the equivalent to mass in the linear motion system. We can think of mass as a resistance to motion (the heavier the mass the lower the acceleration for the same force). The moment of inertia is a resistance to rotation. K 12 mv 2 for the kinetic energy of a mass m moving with speed v. The kinetic We have the formula energy of a point, with mass m, rotating with angular speed at a radius of r about an axis is K 12 mv 2 12 m r 2 12 m 2 r 2 If we have several points, the rotational kinetic energy is the sum of the kinetic energies of the individual points. We will assume that the object is rigid, so all of the points have the same angular speed, . K R Ki i 12 mi ri2 2 i 1 2 and are constant so remove from sum 12 mi ri2 2 i We can define the term in brackets as the objects moment of inertia, I. I mi ri 2 i The units of angular momentum are kg·m2. The fact that we sum over all of the masses illustrates that the moment of inertia of an object is the sum of the moments of inertia of its parts. I Ii i The rotational kinetic energy then becomes K R 12 I 2 Creative Commons Attribution 3.0 Unported License 4 PHYSICS SEMESTER ONE UNIT 7 This has the same form as the linear kinetic energy where the moment of inertia replaces the mass and the angular speed replaces the linear speed. The units of rotational kinetic energy are joules, J. Problem 1 Two balls, both with a mass of 3.5 kg are connected by a mass-less rigid rod. The centre of mass of the system is as the axis of rotation. The balls are 1.0 m from the centre of mass and rotate about it with an angular speed of 12 radians per second. a) Find the moment of inertia of the system. b) Find the rotational kinetic energy of the system. Problem 2 Three 3.6 kg balls are equally space about a circle of radius of 1.0 m forming an equilateral triangle. Find the moment of inertial of the system if a) all of the balls rotate at a radius of 1.0 m about the centre of the system. b) 1.0 m the balls rotate about the bisector of one of the triangle angles. 0.866 m In the second problem, the masses and radii for each are the same for parts a) and b) but the moments of inertia are different. The second problem was chosen to illustrate the fact that the moment of inertia of a system depends on the axis of rotation. Creative Commons Attribution 3.0 Unported License 5 PHYSICS SEMESTER ONE UNIT 7 We can calculate the moment of inertia of any shape by finding the sum of all of the products between mass and the square of the radius for all of the little pieces. I mi ri 2 i or, using calculus, I r 2 dm (FYI only, you are not expected to do integration in this course) For objects with constant density, we can write the derivative of the mass in terms of the density ρ and the volume V. dm dV or dm dV so I r 2 dV Creative Commons Attribution 3.0 Unported License 6 PHYSICS SEMESTER ONE UNIT 7 The following table shows moments of inertia for different objects. If you go into mechanical engineering, you will have a book with many more of these. M is the total mass of the object. Problems on assignments and exams will involve shapes or combinations of shapes from this table. 1. hoop or thin 2. solid sphere R cylinder shell I MR I 52 MR 2 R 2 3. solid cylinder or disk 4. thin spherical shell R I 12 MR2 I 23 MR 2 R 5. long thin rod rotating 6. long thin rod about centre rotating about end I 121 ML2 I 13 ML2 L 7. rectangular plate rotating 8. rectangular plate about centre I 121 M L2 W 2 L rotation about edge L I 13 M L2 W 2 W Creative Commons Attribution 3.0 Unported License L W 7 PHYSICS SEMESTER ONE UNIT 7 Problem 3 For a long thin rod with total mass of 5.0 kg and length 1.2 m, find the moment of inertia a) about an axis of rotation in the centre with the rod perpendicular to the axis, b) about an axis of rotation in the end with the rod perpendicular to the axis. c) Find the moment of inertia of the rod in part b) if there is a ball attached to the rod 1.0 m from the axis. The mass of the ball is 2.0 kg. Part c) of problem 3 illustrates the point that the moment of inertia of an object is the sum of the moments of inertia of the parts. Parallel Axis Theorem With a bit of work, we can show that moment of inertia of any object about an axis through its center of mass is the minimum moment of inertia for any axis in that direction (check text). The moments for shapes 1 through 5 have the axis through the centre of mass of the shape. The parallel axis theorem enables us to expand on the moments of inertia for the shapes noted in the table in the last page. The parallel axis theorem states that the moment of inertia about any axis parallel to that axis through the center of mass is given by I parallel axis ICM Md 2 where d is the distance from the axis through the centre of mass to the parallel axis. axis through centre of mass parallel axis total mass, M centre of mass d There are several physics analyses where we can treat the distributed mass of some object as if all of the mass were concentrated on the centre of mass. This leads to jokes about how physicists can approximate a cow as a single point. The parallel axis theorem indicates that we can account for a shift of an axis parallel to one through the centre of mass by adding the moment of inertia of the entire mass located at the centre of mass to the moment of inertial about the centre of mass axis. The moment of Creative Commons Attribution 3.0 Unported License 8 PHYSICS SEMESTER ONE UNIT 7 inertia about the centre of mass axis accounts for the shape of the object, while the moment of inertia with the concentrated mass at the centre of mass accounts for the location of the actual axis relative to the one through the centre of mass. Problem 4 a) Use the parallel axis theorem to verify that the moment of inertia of a long thin rod rotating about its end is I 1 ML2 given that the moment of inertia for the rod rotating about its centre is 3 1 ML2 . I 12 b) Show that the moment of inertia of a tire (thin walled cylinder) rotating about its edge is I 2ML2 where L is the radius of the tire. TORQUE We have drawn on the similarities between mass and moment of inertia, position and angle, speed and angular speed, and rotational kinetic energy and linear kinetic energy. In rotational systems, we will also talk about the angular acceleration and the torque. Angular acceleration and torque are similar to linear acceleration and force, respectively. Angular acceleration is the rate of change of the angular speed. d dt d 2 dt 2 with the average angular acceleration defined as ave t Both of these have the same form as the linear acceleration. The units of angular acceleration are s-2 or radians per s–2 (radians are considered unitless because they are the ratio of two lengths, the circumference and the diameter). Like the angle and angular velocity the angular acceleration is related to its linear counterpart using a r and a r Creative Commons Attribution 3.0 Unported License 9 PHYSICS SEMESTER ONE The relationship between the net torque net , moment of inertia and angular acceleration is UNIT 7 net I which has the same form as the scalar form of the force in Newton’s second law. Fnet ma Consider the following situation. A force is applied to a wrench, which transfers the force to a nut. F r Fsinθ θ θ Fcosθ d The component of the force that is along the wrench, Fcosθ, does absolutely nothing to turn the nut. It pushes or pulls the nut away from the bolt. The component of the force perpendicular to the wrench, Fsinθ, is the only portion if the force that turns the nut. With this in mind, the torque provided to the nut is the product of the applied force, the distance to the axis of rotation and the sine of the angle between the force and the line joining the place where the force is applied and the axis of rotation. rF sin This torque is the same as that created by the force F applied a distance d from the nut with the force perpendicular to d Fd where d r sin is the moment arm of F. The force is perpendicular to the moment arm. The units of torque are newton-meters, Nm. Dimensionally, 1 Nm is the same as 1 J but we make the distinction because they convey different ideas. In work, W Fd , the force and distance are in the same direction, while in torque, Fd , the force and distances are perpendicular to each other. Creative Commons Attribution 3.0 Unported License 10 PHYSICS SEMESTER ONE UNIT 7 Problem 5 1 A force of 5.0 N is applied to a nut with a moment arm of 0.50 m. A second force of 2.0 N is applied to the nut with moment arm of 1.2 m. Find the total torque on the following system a) assuming the torques produced are in the same direction. b) assuming the torques produced are in the opposite direction. 2 Find the angular acceleration of the nut in the last question if the momentum of inertia is 4.0 kg·m2. a) torques in same direction b) torques in opposite direction Example The following example is a modified version of problem 8 in the dynamics section. In the dynamics problem, the string passed over a frictionless surface so the tension was the same on both sides. In this example the frictionless surface is replaced with a less-than-perfect pulley that has a moment of inertia. The pulley will have some angular acceleration as it moves with the string (no slippage). The angular acceleration is caused by a difference in the torques due to the different tensions on the two sides of the pulley, T1 and T2 . A loonie (mL = 6.99 g) and a dime (mD = 2.09 g) are attached to the ends of a mass-free chain. The chain goes over a pulley (r = 0.10 m, I = 0.00010 kg·m2). Initially motionless, the coins and gear move when released. a) Find the acceleration of the coins. Draw FBDs T1 +y + T2 +y r T1 T2 Pulley FgD FgL Dime Loonie Creative Commons Attribution 3.0 Unported License 11 PHYSICS SEMESTER ONE UNIT 7 The FBDs give us the following. The tensions are tangent to the pulley, perpendicular to the radius to the pulley axis. net I T2 r T1r I a r 1 For the dime, Fnet , D mD a 2 T1 mD g mD a For the loonie, Fnet , L mL a 3 mL g T2 mL a Equations 2 and 3 can be rearranged to give T1 mD a mD g 2' T2 mL g mL a 3' and Substituting these into equation 1 gives T2 r T1r I a r mL g mL a r mD g mD a r I a r Rearranging gives mL gr mL ar mD gr mD ar I mD ar mL ar I a a r a mL gr mD gr r mL gr mD gr I mD r mL r r Creative Commons Attribution 3.0 Unported License 12 PHYSICS SEMESTER ONE UNIT 7 Substituting values (make sure all values are expressed in meters, kilograms and seconds (MKS) or combinations of MKS, like newtons or joules), a 0.00699 9.8 0.10 0.00209 9.8 0.10 0.00209 0.10 0.00699 0.10 0.00010 0.10 2.52 m/s 2 The acceleration of the system is 2.5 m/s2. In the dynamics section problem 8, the acceleration was 5.3 m/s2. The addition of the pulley slows the system down. b) Find the tensions T1 and T2. Substituting into equation 2’ gives T1 mD a mD g 0.00209 kg 2.52 m/s 2 0.00209 kg 9.8 m/s 2 0.0257 N Substituting into equation 3’ gives T2 mL g mL a 0.00699 kg 9.8 m/s 2 0.00699 kg 2.52 m/s 2 0.0509 N The tensions are 0.026 N for the dime side and 0.051 N for the loonie side. In the dynamics section problem 8, the tensions were both 0.032 N. The addition of the not-so-perfect needed non-zero net torque to have acceleration. Problem 6 A 240 kg crate sits on a horizontal surface. The coefficient of kinetic friction between the crate and floor is 0.50. A rope runs horizontally from the crate, over a pulley, to a 175 kg hanging mass. The pulley has a radius of 0.50 m and a moment of inertia of 32 kg·m2. Find the acceleration of the crate. Creative Commons Attribution 3.0 Unported License 13 PHYSICS SEMESTER ONE UNIT 7 WORK AND ROTATIONAL KINETIC ENERGY Consider the small amount work done in rotating an object such that a point P on the object moves by the small arc length displacement ds . F ds dθ r ϕ P O The small amount of work dW is dW F d s F sin rd Here F sin is the tangential component of the force vector F. Recalling the definition of torque, we get dW Fr sin rd d With a bit of work we can get dW Id where the I is constant. Integrating, or adding up all of the little bits of work gives us W f i I d 12 I 2f 12 I i2 K Rf K Ri (once again, you are not expected to use integration on any exams for this course, but it is a good example when you do see integration) The work is equal to the change in rotational kinetic energy (yet another similarity with the linear relations). Creative Commons Attribution 3.0 Unported License 14 PHYSICS SEMESTER ONE UNIT 7 Note that the rotational kinetic energy has the same form as the translational kinetic energy. K R 12 I 2 1 2 mass term velocity term 2 Work-Kinetic Energy Theorem for Rotational Motion The net work done by external forces in rotating a symmetric rigid object about a fixed axis equals the change in the object’s rotation energy. W 12 I 2f 12 Ii2 K Rf K Ri Conservation of energy is also valid for rotational systems We can also show that W which is the angular equivalent of W Fd . You are not responsible for W , but it is yet another relationship that is very similar to a linear equation. Example A fly-wheel is a mechanical device used to store energy. If 2.0106 J of work is done to increase the angular velocity of a fly-wheel from 56 s-1 to 134 s-1, calculate moment of inertia of the fly-wheel. terms: W 2.0 106 J, i 56 s1, f 134 s 1 The work is equal to the change in kinetic energy W K f Ki 12 I 2f 12 I i2 W I 1 2 1 2 2 f i2 I W 2f i2 Creative Commons Attribution 3.0 Unported License 15 PHYSICS SEMESTER ONE UNIT 7 Inserting values I 2.0 106 J 134 s 56 s 1 2 1 2 1 2 269.9 kg·m 2 The moment of inertia of the fly-wheel is 270 kg·m2. Problem 7 A uniform rod of length L = 1.0 m and mass M = 0.30 kg is free to rotate about a frictionless pin through one end. Released from rest in the horizontal position, the rod accelerates due to gravity. a) What is the angular speed when the rod reaches its vertical position? b) Does the angular speed in part a) depend on the mass of the rod? Creative Commons Attribution 3.0 Unported License 16 PHYSICS SEMESTER ONE UNIT 7 ROLLING BODIES An object that is rolling has both linear and rotational kinetic energy. The motion of a tire rolling along the ground can be represented by a combination of both rotational and linear motion. CM is the centre of mass, and the radius of the tire is R. translational motion CM vCM vCM vCM rotational motion v = Rω CM v = Rω combination or translation and rotation motion v = vCM + Rω = 2vCM CM Here we have used the fact that vCM v = vCM – Rω = 0 rolling motion v R vCM Rolling motion is a combination of linear motion and rotation. The kinetic energy of a rolling object is a combination of the linear kinetic energy and the rotational kinetic energy. K KR KL 12 I 2 12 mv 2 Creative Commons Attribution 3.0 Unported License 17 PHYSICS SEMESTER ONE UNIT 7 Example We can use this to find the speed of a can rolling to the bottom of the ramp after starting from rest. The height of the ramp is h. The mass, moment of inertia and radius of the can (thin shelled cylinder) are M, and R. h We will assume we can use the thin shelled cylinder approximation to estimate the moment of inertia of the shell. I MR2 Initial energy is U gi Mgh The final kinetic energy is K f KR KL 12 I 2 12 Mv 2 1 2 1 2 MR 2 M R2 v2 R 2 2 v 2 1 2 Mv R 12 Mv 2 12 Mv 2 12 Mv 2 The rotational kinetic energy equals the linear kinetic energy in this case! Mv 2 Employing the law of conservation of energy (no external force or friction), we can set the initial gravitational energy to the final kinetic energy energies equal to each other. K f U gi M v 2 M gh v gh Creative Commons Attribution 3.0 Unported License 18 PHYSICS SEMESTER ONE UNIT 7 ROTATION DIRECTION If we defined the rotation direction by the direction of the radius vector, we find that the defined direction is constantly changing because the object is rotating. Instead, we can define the direction along the axis of rotation as the rotation direction. The nice thing about this vector is that it doesn’t change as the other vectors rotate about it. For rotation motion in the horizontal plane, we can define the rotation direction as being out of the plane. Counter-clockwise in the horizontal plane is equivalent to up, and clockwise is equivalent to down. With this definition of rotation direction, we can create right-hand rule where the fingers (of the right hand) curl in the motion of the particles in the rotation and the thumb (of the right hand) points in the direction of rotation. The direction notation is useful when looking at the vector products. We will look at some here with torque and angular momentum. In these cases, we can often sketch the situation to determine the direction of the final answer. In the second semester physics courses, the magnetic field interactions are more difficult to imagine so direction definitions are more useful. Example z An object rotates with an angular speed of 2.5 s-1. If the object is rotating counter-clockwise in the x-y plane, what is the direction of rotation (besides counter-clockwise in the x-y plane). y In 2D, we usually draw the 3D x-y-z axes with the z-axis vertical and x-y plane slightly tilted from horizontal (the regular x-y plane can be seen when viewed from above). Applying our right hand rule we see that the x rotation is in the z-direction, 2.5 s-1. The angular velocity is ω 2.5 s1 kˆ . Creative Commons Attribution 3.0 Unported License 19 PHYSICS SEMESTER ONE UNIT 7 VECTOR PRODUCT AND TORQUE Now that we have a convention for the direction of rotation, we can define our torque, angle, angular velocity, angular acceleration, and angular momentum (later) as vectors. We will look primarily at torque. When we looked at energy, we had the “scalar product” or “dot product” of the force and displacement in the definition of work W F d Fx d x Fy d y Fz d z Fd cos where is the angle between the force and displacement. The product, work, is a scalar. There is another product of vectors. This one produces a vector result and is known as the vector product. The vector product (also called the “cross product”) is defined as follows A B Ay Bz Az By ˆi Az Bx Ax Bz ˆj Ax By Ay Bx kˆ This has some interesting properties B A B y Az Bz Ay ˆi Bz Ax Bx Az ˆj Bx Ay B y Ax kˆ Ay Bz Az B y ˆi Az Bx Ax Bz ˆj Ax B y Ay Bx kˆ A B From here, we can get A A A A ( A A , the colour is used to show the position swap) The only value that can equal it negative is 0 so A A 0 We can also use what is known as the determinant of a matrix to find the cross product. A matrix is a rectangular array of numbers or variables. Examples of a 2 by 2 matrix K, a 3 by 3 matrix L, and a 3 by 2 matrix M are shown below. 1 2 K , 4 3 a c L d e g h b f , i Creative Commons Attribution 3.0 Unported License j k M 5 6 8 7 20 PHYSICS SEMESTER ONE UNIT 7 For the cross product using the determinant (symbolized by replacing [ ] with | |), place the direction vectors in the top row, the first vector in the middle row, and the second vector in the bottom row. ˆi ˆj kˆ AB Ax Ay Az Bx By Bz To find the î components (x-direction): 1. Block out the row and column containing î . ˆi 2. ˆj kˆ AB Ax Ay Az Bx By Bz Match the unblocked terms along the diagonals. ˆi 3. ˆj kˆ AB Ax Ay Az Bx By Bz Ay Bz and Az B y The positive product is the one with the second row term to the right of the blocked off column. If there is no column to the right, use the far left column. A B Ay Bz Az By ˆi Az Bx Ax Bz ˆj Ax By Ay Bx kˆ 4. Repeat for the other directions. For the ĵ direction (y-direction) ˆi ˆj kˆ AB Ax Ay Az Bx By Bz A B Ay Bz Az By ˆi Az Bx Ax Bz ˆj Ax By Ay Bx kˆ Creative Commons Attribution 3.0 Unported License 21 PHYSICS SEMESTER ONE UNIT 7 For the k̂ direction (z-direction) ˆi ˆj kˆ AB Ax Ay Az Bx By Bz A B Ay Bz Az By ˆi Az Bx Ax Bz ˆj Ax By Ay Bx kˆ Right Hand Rule II Taking the right hand, if we extend the fingers in the direction of the first vector (vector A). Then curl them towards the second vector (vector B) (alternately, the line coming out of the palm is in the direction of B). The direction of the extended thumb is in the direction of the cross product A × B. Example The cross product of A 1 ˆi with B 1 ˆj , A B Ay Bz Az By ˆi Az Bx Ax Bz ˆj Ax By Ay Bx kˆ 0 0 0 1 ˆi 0 0 1 0 ˆj 1 1 0 0 kˆ 1 kˆ If we have two vectors, we can always find a plane that has both vectors A 1 ˆi with B 1 ˆj are both in the x-y plane. A×B B The cross product is always perpendicular to the plane of the two vectors in the cross product. A If we measure the angle between the two vectors in the plane, the magnitude of the cross product is the product of the magnitude of the two vectors times the |A × B| = AB sin sine of the angle between them. Creative Commons Attribution 3.0 Unported License B Aφ 22 PHYSICS SEMESTER ONE UNIT 7 For the torque vector, we have τ rF ry Fz rz Fy ˆi rz Fx rx Fz ˆj rx Fy ry Fx kˆ For force and radius in two dimensions (in x-y plane), this simplifies greatly to τ rF rx Fy ry Fx kˆ ry 0 0 Fy ˆi 0 Fx rx 0 ˆj rx Fy ry Fx kˆ rF sin kˆ Where is the angle between the radius and force vectors, measured from the radius vector. Examples A force of F 8.66 N ˆi 5.00 N ˆj is applied to an object r 1.00 m ˆi 1.73 m ˆj from the axis of rotation. a) Find the torque using the cross product. τ rF 1.73 0 0 5.00 ˆi 0 8.66 1.00 0 ˆj 1.00 5.00 1.73 8.66 Nm kˆ 5.00 15.00 Nm kˆ 10.0 Nm kˆ b) Find the magnitude of the torque using the magnitudes of the force and radius, and the sine of the angle from the radius vector to the force vector, -30°. F 5.002 8.662 10.0 N , r 1.732 1.002 2.00 m rF sin 2.00 m 10.0 N sin 30 10.0 Nm Creative Commons Attribution 3.0 Unported License 23 PHYSICS SEMESTER ONE UNIT 7 Your choice of method for calculating the torque depends on the form of the original values. Example b) is easier when you have the magnitude-direction form of the vectors, while example a) is easier with the component form vectors. Alternate Direction Rules (please, ignore if confusing) Please note, if you insist on using your left hand, you can paint (draw, tattoo …) an arrow on your thumb pointing towards your palm, to point in the cross product direction. Alternatively, you can use the opposite of the direction indicated by the left thumb (not the arrow), or multiply the direction indicated by the left thumb by -1. It is usually easier to just use your right hand. If you imagine a clock showing 3:00, you can think of the fist vector as the minute hand pointing at the 12 (XII) and the second vector along the hour hand pointing at the 3 (III). The direction of the cross product is into the face of the clock. Creative Commons Attribution 3.0 Unported License 24 PHYSICS SEMESTER ONE UNIT 7 Problem 8 Find the torque for the following a) F 2.6 N ˆi 5.0 N ˆj, r 1.8 m ˆi 1.1 m ˆj b) F = 12.5 N, r = 3.4 m, 85 , magnitude only c) F = 12.5 N, r = 3.4 m, 5 , magnitude only Creative Commons Attribution 3.0 Unported License 25 PHYSICS SEMESTER ONE UNIT 7 ANGULAR MOMENTUM Going back to the affect of torque, we had the relationship net I Recalling the definition of the angular acceleration net I I I t f i t I f I i t It is convenient to introduce the angular momentum L as the product of the moment of inertia and the angular speed. L I This is the counterpart to the linear momentum. For a given speed, an object with a large mass will have a higher momentum than object with a small mass. For a given angular speed, an object with a large moment of inertia will have a larger angular momentum than an object with a small moment of inertia. The units of angular momentum are kg·m2/s. We can rewrite the average net torque on an object as net , ave L f Li t L t Using calculus, we can use this to find the instantaneous net torque as net dL dt The net torque acting on the system is equal to the time rate of change of the object’s angular momentum. Creative Commons Attribution 3.0 Unported License 26 PHYSICS SEMESTER ONE UNIT 7 As with linear momentum and energy, angular momentum is conserved under specific conditions. Angular momentum is conserved when there is no net torque on the system. If net 0 then L f Li . The classic demonstration of this is the spinning stool demonstration (a similar, less artistic, version occurs in figure skating and dancing). A person sits on a stool with weights extended out from his/her body. Using whatever method they have, they start rotating. Keeping the feet off the ground, the person will continue to rotate and a fairly constant speed (slowing a bit due to friction). By pulling their arms in their angular speed can be increased greatly. Alternately, a person with a high rotation can slow by spreading their arms and legs out. Once started, the person is rotating without external torques. Conservation of angular momentum tells us that the angular momentum is constant. If net 0 then L f Li or I f f I ii . When the person is spread out they have a large moment of inertia. When the person tucks in, their moment of inertia decreases. Their angular momentum must increase to keep the angular momentum constant. f Ii i 1i If In figure skating, a skater exploits this property when they spin. Dancers can leap while spinning, then extend their arms and appear to freeze in mid air before tucking in and speeding up as they land. Problem 9 A skater, with a moment of inertia of 9.0 kg·m2, is rotating at 3.0 radians per second with arms extended. a) Find the skater’s angular momentum. b) Find the skater’s angular speed if she pulls her arms and legs in, reducing her moment if inertia to 3.2 kg·m2. Creative Commons Attribution 3.0 Unported License 27 PHYSICS SEMESTER ONE UNIT 7 The vector form of the angular momentum is expressed by the equation L rp Like the translational momentum, angular momentum is actually a vector, and like torque, we use the cross product to find the angular momentum vector. Here we used the fact that dL τ net dt r Fnet r dp dt To be truly consistent with the product rule in calculus, we really need dL dp dr r p , dt dt dt but we will considering cases where r as constant. Note that we don’t need an actual rotating object with this definition. Consider the following example: A 920 kg car is driving along a straight highway with a velocity of v 32 m/s ˆi . Bob is located 15 m at an angle perpendicular to the highway. The car a 100 m to Bob’s left. Find the moment of inertia of the car relative to Bob (axis). v 32 m/s ˆi r 120 m ˆi 15 m ˆj We can use the sketch to show that r 120 m ˆi 15 m ˆj . Noting that p mv , the angular momentum is L rp 120 m ˆi 15 m ˆj 920 kg 32 m/s ˆi 4.4 105 kg m 2 /s kˆ The angular momentum is 4.4 105 kg m 2 /s kˆ or 4.4105 kg·m2/s in the downwards direction. Creative Commons Attribution 3.0 Unported License 28 PHYSICS SEMESTER ONE UNIT 7 We are usually just interested in the magnitude. If we know the magnitudes of and angle between the position and momentum vectors, L rp sin rmv sin Example A 920 kg car is driving along a straight highway 32 m/s. Bob is located 121 m from car at an angle of 7.1° from the highway. Find the moment of inertia of the car relative to Bob (axis). terms: m 920 kg, v 32 m/s, r 121 m, 7.1, L ? equation: L rmv sin L 121 m 920 kg 32 m/s sin 7.1 calculate: 440300 kg m2 s The car has an angular momentum of 4.4105 kg·m2/s relative to Bob. Like the moment arm with torque, we really only need to consider the radius perpendicular to the momentum because r sin r so L r p r mv In the last example, the radius perpendicular to the direction of the car is r r sin 121 m sin 7.1 15.0 m keep extra decimal place for intermediate term The angular momentum is then L r mv 15.0 m 920 kg 32 m/s 4.4 105 kg·m 2 /s Creative Commons Attribution 3.0 Unported License 29 PHYSICS SEMESTER ONE UNIT 7 CENTRE OF GRAVITY A few sections ago we had the centre of mass of an object defined as xCM m x m x M m i i i i yCM i m y m y M m i i i i i We will consider a two dimensional object. The arguments can be easily extended to 3D. Gravity acts on the all of the individual masses in a body. We can define a centre of gravity xCG using the following equation for the total torque acting on the object due to gravity. net , g gMxCG Gravitational forces are parallel to the y-axis so only the x-components of each position vector contribute the net torque. The torque for an the ith individual mass is xi τ i ri Fi ri Fi sin i yi xi mi g ri Fgi = mig The net torque on the object is the sum of all of the gravitational forces on all of the particles net , g i mi xi g If the gravitational field is constant within the object (good assumption for most things under 30 km in diameter), the net gravitational torque simplifies to net , g g mi xi gM mi xi M gMxCM When the gravitational field is constant over the entire object, the centre of gravity is equal to the centre of mass. Creative Commons Attribution 3.0 Unported License 30 PHYSICS SEMESTER ONE UNIT 7 Problems 10 A torque of 13 Nm is applied to an object for 4.2 seconds, increasing its angular speed from 1.1 radians per second to 3.4 radians per second. Calculate the moment of inertia of the object. You can now revisit the PHeT torque and rotational motion applet to look at the relationships between the angular acceleration, torque and moment of inertia for the lady bug merry-go-round, http://phet.colorado.edu/new/simulations/sims.php?sim=Torque. In the torque section, you can vary the applied force and radius of force (moment arm) to examine their effects on the torque. Unfortunately, there are no examples where you can vary the angle between the force and radius. In the moment of inertia section, get the disk rotating by click-dragging the mouse on it. You can then vary the moment of inertia by altering the inner and outer diameters of the disk. Notice how the angular speed speeds up/slows down when the moment of inertia is reduced/increased. For some reason, the angular acceleration does not register this change in angular speed. In the angular momentum section, check out how the angular moment stays constant while the angular speed changes when you alter the moment of inertia. Creative Commons Attribution 3.0 Unported License 31 PHYSICS SEMESTER ONE UNIT 7 STATIC EQUILIBRIUM static – lacking change equilibrium – state in which opposing forces & torques are balanced A system is in static equilibrium if the net external force and net external torque are zero. Fnet 0 τ net 0 These sums combine with Newton’s Laws to give us the conditions where the linear acceleration and angular acceleration must be zero. a0 0 The direction for the angular acceleration is not critical for this course. We can often define directions with respect to the pivot points in a sketch. Torques that would cause rotations in one direction are positive, the others are negative. In equilibrium questions, the sum of the torques on an object is zero so there is really no rotation anyway. Creative Commons Attribution 3.0 Unported License 32 PHYSICS SEMESTER ONE UNIT 7 Example Dr. Evil has devised an “escape-proof” plan for eliminating Anne and Bob. Anne, mass 61 kg, is dropped on one end of a 9.0 m see-saw balanced above a pool of killer goldfish. Bob, mass 79 kg, is dropped on the other side 2.3 m from the axis of the see-saw. a) How far must Anne be from the axis to ensure that she and Bob are in static equilibrium, with neither dipping into the pool. mB = 79 kg mA = 61 kg ? 2.3 m 9.0 m Let’s draw the FBD for the see-saw. mS-Sg mAg xA FN 2.3 m mBg In this type of FBD, we account for the fact that the forces on the see-saw are at different locations. The locations of the forces are included because they affect the torque on the see-saw. Here, the unknown variables are mss , FN and x A . It turns out that we can ignore mss and FN if we consider the see-saw pivot axis for our rotation (non-rotation). The net torque about the axis is net Fi xi mB g 2.3 m mA gxA mss g 0 FN 0 For static equilibrium net 0 Creative Commons Attribution 3.0 Unported License 33 PHYSICS SEMESTER ONE UNIT 7 Solving for xA, 0 mB g 2.3 m mA gx A mA g x A mB g 2.3 m xA mB 2.3 m mA 79 kg 2.3 m 61 kg 2.98 m Anne is at -3.0 m, 3.0 m on the side of the see-saw opposite Bob. b) If Anne can reach the end of her side of the see-saw, she can reach a switch that locks the seesaw in place. If Bob moves toward his end at a rate of 1.0 m/s, how fast must Anne go in order to ensure that the see-saw is balanced while they move to their ends? Bob’s position, in m, at time t is xB 2.3 1.0t Anne’s position, in m, at time t is xA 3.0 vAt At each location, the positions of Anne and Bob must satisfy the equilibrium net torque condition. mB gxB mA gxA 0 Canceling out gravity and inserting other terms, we have 79 2.3 1.0t 61 2.98 vAt 0 181.7 79t 181.7 61v At 0 79t 61v At 0 Solving for Anne’s speed vA 79t 61t 1.295 m/s Anne must move at about 1.3 m/s away from Bob to ensure static equilibrium is maintained. Creative Commons Attribution 3.0 Unported License 34 PHYSICS SEMESTER ONE UNIT 7 Equilibrium Problem Solving Strategy 1. Draw a sketch. Include coordinates and a convenient rotation axis. The choice of the rotation axis is arbitrary, though some choices are better than others. The choice is arbitrary because the object in equilibrium is not rotating about any axis. Whichever axis you select, the object is not rotating about it. A convenient rotation axis is usually one that eliminates the most unknown terms. In the last example, choosing the rotation axis at the pivot for the see-saw allowed us to ignore the normal force and weight of the see-saw. 2. Draw a FBD. Show all external forces and distances. If there is more than one object, draw separate diagrams. 3. Apply net 0 . This generates an equation relating the forces and their positions. It helps to note the direction or expected direction of each torque on the FBD. This way you can account for the torque signs within the equation. 4. Apply Fnet 0 . Apply this for all directions, ignoring the positions of the forces. This yields an equation linking the forces for each direction. 5. Solve. We have two or three equations. Solve the system of equations for unknown variables. Creative Commons Attribution 3.0 Unported License 35 PHYSICS SEMESTER ONE UNIT 7 Example A sign hangs from a beam outside Bob’s Butcher shop. The beam is 2.5 m long and weighs 32 N with a centre of gravity 1.0 m from the wall. The beam’s centre of mass is the centre of the beam. The 85 N sign is suspended from a point 1.5 m from the wall. A cable extends from the end of the beam to the wall above the cable. The cable makes an angle of 37° with the beam. Find the tension in the cable and the force applied by the wall in the x- and y-directions. Terms: 1. xT 2.5 m, T ?, 37, xB 1.0 m, FgB 32 N, xS 1.5 m, FgS 85 N, FWx ?, FWy ? Sketch y x θ = 37° Bob’s axis of rotation 2. Fine Meats FBD for the beam T Fw 37° FgB FgS 1.0 m 1.5 m 2.5 m Creative Commons Attribution 3.0 Unported License 36 PHYSICS SEMESTER ONE UNIT 7 Here, we have three unknowns are the x and y-components of the wall force and the magnitude of the tension. The arrow indicating the force from the wall is only a guess. You can get a rough idea of the direction and size knowing that it must balance the other forces. 3. In the sketch, the axis of rotation is the place where the wall meets the beam. This choice means that the radius for the wall force (and the wall torque) is zero. The system is in static equilibrium so the total torque on the beam gives us net FWy 0 FgB xB FgS xS TxT sin T 0 0 32 N 1.0 m 85 N 1.5 m T 2.5 m sin 37 Solving for T 1.505 m T 32 Nm 127.5 Nm T 159.5 Nm 1.505 m 106 N The tension on the wire is 110 N. This stage doesn’t always simplify to a single value. Sometimes the result is an equation with two or three variables. 4. The big trick here is to ignore all of the distances. As far as the Fnet 0 analysis is concerned, the FBD for the beam is T Fw Ty 37° Fwy Tx Fwx FgB FgS Creative Commons Attribution 3.0 Unported License 37 PHYSICS SEMESTER ONE UNIT 7 Looking at the forces in the x-direction, we have Fnet , x 0 FWx Tx 0 FWx Tx T cos T 106 N cos 37 84.67 N In the y-direction Fnet , y 0 FWy FgB FgS Ty 0 Solving for FWy FWy FgB FgS Ty 32 N 85 N T sin T 119 N 106 N sin 37 55.2 N The tension is 110 N and the force from the wall is 85 N in the x-direction and 55 N in the ydirection. Creative Commons Attribution 3.0 Unported License 38 PHYSICS SEMESTER ONE UNIT 7 Problem 11 A two dimensional tractor sits on level ground. The distance between the front and back wheel is 2.7 m. The 1.0 x 104 N weight of the tractor has a centre of mass 1.0 m from the back of the tractor. Calculate the weight on each wheel (two variables). Problem 12 A 3.0 m ladder leans against a wall, making an angle of 51° with the ground. Fiction between the ladder and ground keeps the ladder from slipping. There is no friction between the ladder and the wall so the wall force is perpendicular to the wall. Calculate the normal force on the ladder, the friction force and the force from the wall. Check out equilibrium problems in your textbook. Creative Commons Attribution 3.0 Unported License 39 PHYSICS SEMESTER ONE UNIT 7 SUMMARY OF ROTATIONAL MOTION AND COMPARISON WITH LINEAR MOTION Rotational Motion Angle Link Linear Motion Position, Distance, Arc Length x, s s r s on curve θ r d dt Angular Speed Speed v dx vr Angular Acceleration dt v perpendicular to r θ r d dt Acceleration a dv ar dt a perpendicular to r θ r If is constant If a is constant f i t v f vi at f i it 12 t 2 x f xi vit 12 at 2 2f i2 2 f i v2f vi2 2ax f xi Moment of Inertia I Mass m Torque Fr sin Fd , d is the moment arm Force F Net Torque net I Work W f Net Force Fnet ma d or W Work W i xf x Fdx or i W F x Fd Rotational Kinetic Energy K R 1 I 2 2 Kinetic Energy K 1 mv 2 2 Power P Power P Fv Angular Momentum L I Net Torque II net dL dt Creative Commons Attribution 3.0 Unported License Momentum p mv Net Force II Fnet dp dt 40 PHYSICS SEMESTER ONE UNIT 7 NANSLO Physics Core Units and Laboratory Experiments by the North American Network of Science Labs Online, a collaboration between WICHE, CCCS, and BCcampus is licensed under a Creative Commons Attribution 3.0 Unported License; based on a work at rwsl.nic.bc.ca. Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges. Creative Commons Attribution 3.0 Unported License 41