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PHYS 1443 – Section 001 Lecture #18 Monday, April 18, 2011 Dr. Jaehoon Yu • • • • • • Torque & Vector product Moment of Inertia Calculation of Moment of Inertia Parallel Axis Theorem Torque and Angular Acceleration Rolling Motion and Rotational Kinetic Energy Today’s homework is homework #10, due 10pm, Tuesday, Apr. 26!! Announcements • Quiz Wednesday, Apr. 20 – Beginning of the class – Covers from CH9.5 to CH10 • Reading assignment: CH10 – 10. • Remember to submit your extra credit special project Monday, April 18, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 2 Torque Torque is the tendency of a force to rotate an object about an axis. Torque, , is a vector quantity. F r P The line of Action d2 d Moment arm F2 Consider an object pivoting about the point P by the force F being exerted at a distance r from P. The line that extends out of the tail of the force vector is called the line of action. The perpendicular distance from the pivoting point P to the line of action is called the moment arm. Magnitude of torque is defined as the product of the force exerted on the object to rotate it and the moment arm. When there are more than one force being exerted on certain points of the object, one can sum up the torque generated by each force vectorially. The convention for sign of the torque is positive if rotation is in counter-clockwise and negative if clockwise. Monday, April 18, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu rF sin Fd 1 2 F1d1 F2d2 3 Ex. 10 – 7: Torque A one piece cylinder is shaped as in the figure with core section protruding from the larger drum. The cylinder is free to rotate around the central axis shown in the picture. A rope wrapped around the drum whose radius is R1 exerts force F1 to the right on the cylinder, and another force exerts F2 on the core whose radius is R2 downward on the cylinder. A) What is the net torque acting on the cylinder about the rotation axis? The torque due to F1 R1 R2 1 R1F1 and due to F2 So the total torque acting on the system by the forces is 1 2 R2 F2 2 R1F1 R2 F2 Suppose F1=5.0 N, R1=1.0 m, F2= 15.0 N, and R2=0.50 m. What is the net torque about the rotation axis and which way does the cylinder rotate from the rest? Using the above result Monday, April 18, 2011 R1F1 R2 F2 5.0 1.0 15.0 0.50 2.5 N m PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu The cylinder rotates in counter-clockwise. 4 Torque and Vector Product z Let’s consider a disk fixed onto the origin O and the force F exerts on the point p. What happens? rxF O r The disk will start rotating counter clockwise about the Z axis y The magnitude of torque given to the disk by the force F is p θ Fr sin F x But torque is a vector quantity, what is the direction? How is torque expressed mathematically? What is the direction? The direction of the torque follows the right-hand rule!! The above operation is called the Vector product or Cross product What is the result of a vector product? Another vector Monday, April 18, 2011 rF C A B ur ur ur ur ur C A B A B sin What is another vector operation we’ve learned? ur ur ur ur Scalar product C A B A B cos PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu Result? A scalar 5 Properties of Vector Product Vector Product is Non-commutative What does this mean? ur ur ur ur If the order of operation changes the result changes A B B A ur Following the right-hand rule, the direction changes ur ur Vector Product of two parallel vectors is 0. ur A B B A ur ur ur ur ur ur ur C A B A B sin A B sin 0 0 Thus, ur ur AA0 If two vectors are perpendicular to each other ur ur ur ur ur ur ur ur o A B A B sin A B sin 90 A B AB Vector product follows distribution law ur ur ur ur ur ur ur A BC A B AC The derivative of a Vector product with respect to a scalar variable is ur ur ur ur u r u r d A B dA dB B A dt dt dt Monday, April 18, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 6 More Properties of Vector Product The relationship r r between r unit vectors, i, j and k r r r r r r i i j j k k 0 r r r r r i j j i k r r r r r j k k j i r r r r r k i i k j Vector product of two vectors can be expressed in the following determinant form ur ur AB r i r j r k Ax Ay Az Bx By Bz Ay Bz Az By Monday, April 18, 2011 r i Ay Az By Bz r j Ax Az Bx Bz r k Ax Ay Bx By r r r i Ax Bz Az Bx j Ax By Ay Bx k PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 7 Moment of Inertia Rotational Inertia: For a group of particles Measure of resistance of an object to changes in its rotational motion. Equivalent to mass in linear motion. I mi ri2 i What are the dimension and unit of Moment of Inertia? For a rigid body I r 2 dm ML kg m 2 2 Determining Moment of Inertia is extremely important for computing equilibrium of a rigid body, such as a building. Dependent on the axis of rotation!!! Monday, April 18, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 8 Example for Moment of Inertia In a system of four small spheres as shown in the figure, assuming the radii are negligible and the rods connecting the particles are massless, compute the moment of inertia and the rotational kinetic energy when the system rotates about the y-axis at angular speed ω. y m Since the rotation is about y axis, the moment of inertia about y axis, Iy, is b l M O l 2 2 2 2 2 2 I m 0 m 0 m r 2Ml Ml Ml ii x M b m i This is because the rotation is done about y axis, and the radii of the spheres are negligible. 1 2 1 K R I 2Ml 2 2 Ml 2 2 2 2 Why are some 0s? Thus, the rotational kinetic energy is Find the moment of inertia and rotational kinetic energy when the system rotates on the x-y plane about the z-axis that goes through the origin O. 2 2 2 I mi ri 2 Ml Ml 2 mb2 mb 2 2 Ml mb i Monday, April 18, 2011 1 1 K R I 2 2Ml 2 2mb2 2 Ml 2 mb2 2 2 2 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 9 Calculation of Moments of Inertia Moments of inertia for large objects can be computed, if we assume the object consists of small volume elements with mass, mi. 2 2 I lim r m r i i The moment of inertia for the large rigid object is dm m 0 i i It is sometimes easier to compute moments of inertia in terms How can we do this? of volume of the elements rather than their mass Using the volume density, , replace dm dm dV The moments of 2 I r dV dV dm in the above equation with dV. inertia becomes Example: Find the moment of inertia of a uniform hoop of mass M and radius R about an axis perpendicular to the plane of the hoop and passing through its center. y O The moment of inertia is dm R Monday, April 18, 2011 x What do you notice from this result? I r 2 dm R2 dm MR 2 The moment of inertia for this object is the same as that of a point of mass M at the distance R. PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 10 Ex.10 – 11 Rigid Body Moment of Inertia Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an axis perpendicular to the rod and passing through its center of mass. M The line density of the rod is y L M so the masslet is dm dx dx L dx x x L The moment of inertia is M L L 2 3L 2 3 What is the moment of inertia when the rotational axis is at one end of the rod. Will this be the same as the above. Why or why not? Monday, April 18, 2011 L /2 M 1 3 x2 M x r dm I dx L /2 L 3 L / 2 L L /2 2 I 3 M L3 ML2 4 3L 12 L M 1 3 x2 M 2 x dx r dm 0 L 3 0 L 2 M M 3 ML 3 L 0 L 3L 3L 3 L Since the moment of inertia is resistance to motion, it makes perfect sense for it to be harder to move when it is rotating about the axis at one end. PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 11 Parallel Axis Theorem Moments of inertia for highly symmetric object is easy to compute if the rotational axis is the same as the axis of symmetry. However if the axis of rotation does not coincide with axis of symmetry, the calculation can still be done in a simple manner using parallel-axis theorem. I I CM MD 2 y Moment of inertia is defined as r yCM y y’ (x,y) D xCM What does this theorem tell you? Monday, April 18, 2011 x xCM x ' y yCM y' One can substitute x and y in Eq. 1 to obtain 2 2 I xCM x' yCM y' dm CM 2 2 2 2 (xCM,yCM) xCM yCM dm 2x x'dm 2y y'dm x' y' dm CM CM x’ x Since x and y are I r 2 dm x 2 y 2 dm (1) Since the x’ and y’ are the x' dm 0 y' dm 0 x distance from CM, by definition Therefore, the parallel-axis theorem 2 2 I xCM yCM dm x'2 y'2 dm MD2 ICM Moment of inertia of any object about any arbitrary axis are the same as the sum ofPHYS moment of inertia 1443-001, Spring for 2011a rotation about the CM and that 12of Dr. Jaehoon Yu axis. the CM about the rotation Example for Parallel Axis Theorem Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an axis that goes through one end of the rod, using parallel-axis theorem. The line density of the rod is y so the masslet is CM dm dx dx x L x The moment of inertia about the CM M L M dx L L /2 I CM M 1 3 x2 M 2 x r dm dx L /2 L 3 L / 2 L L /2 3 3 M L L M L3 ML2 2 3L 4 3L 2 12 2 Using the parallel axis theorem I I CM ML2 L ML2 ML2 ML2 2 D M M 12 2 12 4 3 The result is the same as using the definition of moment of inertia. Parallel-axis theorem is useful to compute moment of inertia of a rotation of a rigid object with complicated shape about an arbitrary axis Monday, April 18, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 13 Check out Figure 10 – 20 for moment of inertia for various shaped objects Monday, April 18, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 14 Torque & Angular Acceleration Ft r F r Let’s consider a point object with mass m rotating on a circle. What forces do you see in this motion? m The tangential force Ft and the radial force Fr Ft mat mr The torque due to tangential force Ft is F r ma r mr 2 I t t The tangential force Ft is What do you see from the above relationship? What does this mean? I Torque acting on a particle is proportional to the angular acceleration. What law do you see from this relationship? Analogs to Newton’s 2nd law of motion in rotation. How about a rigid object? The external tangential force δFt is d Ft d mat d mr dFt 2 r dm d F r d The torque due to tangential force F is t t δm The total torque is lim d d lim r 2d m r 2 dm I d 0 d m0 r Contribution from radial force is 0, because its What is the contribution due line of action passes through the pivoting O to radial force and why? Monday, April 18, 2011 PHYS 1443-001, Spring 2011point, making the moment arm 0. 15 Dr. Jaehoon Yu Rolling Motion of a Rigid Body What is a rolling motion? A more generalized case of a motion where the rotational axis moves together with an object A rotational motion about a moving axis To simplify the discussion, let’s make a few assumptions 1. 2. Limit our discussion on very symmetric objects, such as cylinders, spheres, etc The object rolls on a flat surface Let’s consider a cylinder rolling on a flat surface, without slipping. Under what condition does this “Pure Rolling” happen? The total linear distance the CM of the cylinder moved is sR R s s=Rθ Monday, April 18, 2011 Thus the linear speed of the CM is ds d vCM R R dt dt The condition for a “Pure Rolling motion” PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 16 More Rolling Motion of a Rigid Body The magnitude of the linear acceleration of the CM is P’ CM dvCM d aCM R R dt dt As we learned in rotational motion, all points in a rigid body 2vCM moves at the same angular speed but at different linear speeds. vCM CM is moving at the same speed at all times. At any given time, the point that comes to P has 0 linear speed while the point at P’ has twice the speed of CM P Why?? A rolling motion can be interpreted as the sum of Translation and Rotation P’ CM P vCM P’ vCM CM v=0 vCM Monday, April 18, 2011 + v=Rω v=Rω 2vCM P’ = P PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu CM vCM P 17 Kinetic Energy of a Rolling Sphere Let’s consider a sphere with radius R rolling down the hill without slipping. R h θ vCM Since vCM=Rω What is the speed of the CM in terms of known quantities and how do you find this out? Monday, April 18, 2011 1 2 1 R22 K ICM M 2 2 2 1 2 1 v C M ICM MvCM 2 R 2 1 I CM 2 2 M v CM 2 R Since the kinetic energy at the bottom of the hill must be equal to the potential energy at the top of the hill 1 I CM 2 Mgh M v 2 K CM 2 R PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 2 gh v CM 2 1 I /MR CM 18 Ex. 10 – 16: Rolling Kinetic Energy For solid sphere as shown in the figure, calculate the linear speed of the CM at the bottom of the hill and the magnitude of linear acceleration of the CM. Solve this problem using Newton’s second law, the dynamic method. What are the forces involved in this motion? Gravitational Force, Frictional Force, Normal Force Newton’s second law applied to the CM gives n f M h Mg Mg sin fMa F n Mg cos 0 F θ x CM y Since the forces Mg and n go through the CM, their moment arm is 0 fRICM and do not contribute to torque, while the static friction f causes torque CM We know that 2 2 ICM MR 5 a R CM Monday, April 18, 2011 We obtain 2 2 MR ICM aCM 2Ma f 5 CM R R R 5 Substituting f in dynamic equations 5 7 a g sin Mg sin Ma CM CM 7 5 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 19