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ROTATIONAL MOTION AND EQUILIBRIUM Angular Quantities of Rotational Motion Rotational Kinematics Torque Center of Gravity Moment of Inertia Rotational Kinetic Energy Angular Momentum Rotational Equilibrium ANGULAR SPEED AND ANGULAR ACCELERATION: w = angular speed rad/s deg/s rev/s a = angular acceleration rad/s2 deg/s2 rev/s2 q rotating drum axis Dq w= Dt Dw a= Dt ANGULAR CONVERSIONS: 1 rev = 2p rad = 360 deg p rad = 180 deg Convert 246o to radians. Convert 16.4 rev to degrees. Convert 246o to radians. 246 = 246 o o p 180 o = 1.37p rad Convert 16.4 rev to degrees. 360o 16.4rev = 16.4rev = 5904o rev ROTATIONAL KINEMATICS: a =a Uniform angular acceleration: q = instantaneous angular position w = instantaneous angular speed q = q o wo a t 1 2 2 w = wo a t w = w 2a q q o 2 2 o EXAMPLE: Through what angle does a wheel rotate if it begins from rest and accelerates at 3 o/s2 for 15 seconds? Also calculate its final angular speed. ANSWER: Through what angle does a wheel rotate if it begins from rest and accelerates at 3 o/s2 for 15 seconds? Also calculate its final angular speed. q = q o w ot a t 1 2 q = 00 1 2 2 3 15s deg s2 q = 337 deg w = wo a t w = 0 3 deg s 15s 2 w = 45deg/ s 2 ROTATIONAL AND LINEAR QUANTITIES: w r s = rq v = rw a = ra Angular quantities must be expressed in terms of radians. EXAMPLE: A point on the rim 5cm from the axis of rotation moves through an angle of 15 degrees. How far has it traveled? axis ANSWER: A point on the rim 5cm from the axis of rotation moves through an angle of 15 degrees. How far has it traveled? p s = 0.05m 15 = 0.13m o 180 o EXAMPLE: A wheel of radius r is rolling along the ground without slipping so that its center moves with a constant speed v? What are the speeds of the points at its top and at its bottom? Since it is not slipping the speed at the center must be v = 2pr/T. The rotational speed must be w = 2p/T. The speed of a point on the rim, relative to the center, must be rw = 2pr/T. This is equal to v. At its bottom the rim is moving backwards relative to the center so that its velocity relative to the ground if v -v = 0. At its top the rim is moving forward relative to the center to that its 2v velocity is v + v = 2v. v TORQUE: Torque, t, is produced when a force acts on a body along a line that is displaced from a pivot point or axis of rotation. LINE OF ACTION: The line of action of a force is the line along which the force acts. LEVER ARM: The lever arm of a force is the distance from the pivot or axis of rotation to the point where the force is applied. Bar t = Fr F = F sin q t = F sin q r r pivot F q F Force line of action The torque produced by this force is trying to rotate the bar in a counter-clockwise direction. HOW TO CALCULATE A TORQUE: 1. In a sketch, locate the pivot point or axis of rotation. 2. Locate the force producing the torque and where it is applied. 3. Determine the length of the line segment from the pivot to the point where the force is applied. This is the lever arm. 4. Determine the angle between the force and the lever arm. The force multiplied by the sine of this angle is the force component perpendicular to the lever arm. 5. The magnitude of the torque is the product of the force perpendicular component and the lever arm. 6. Determine the direction of rotation the torque is trying to induce as being either clockwise or counter-clockwise. EXAMPLE: Calculate the torque produced by the force acting on the rod as shown. 1.25m pivot 57o F = 16N EXAMPLE: Calculate the torque produced by the force acting on the rod as shown. 1. Identify the pivot (left end of rod) 1.25m pivot 57o F = 16N EXAMPLE: Calculate the torque produced by the force acting on the rod as shown. 1. Identify the pivot (left end of rod) 1.25m pivot 57o 2. Identify the force ( F = 16 N) F = 16N EXAMPLE: Calculate the torque produced by the force acting on the rod as shown. 1. Identify the pivot (left end of rod) 1.25m pivot 57o 2. Identify the force ( F = 16 N) 3. Determine the lever arm ( r = 1.25 m) F = 16N EXAMPLE: Calculate the torque produced by the force acting on the rod as shown. 1. Identify the pivot (left end of rod) 1.25m pivot 57o 2. Identify the force ( F = 16 N) 3. Determine the lever arm ( r = 1.25 m) 4. Determine the angle between the force and the lever arm (q = 57o) F = 16N EXAMPLE: Calculate the torque produced by the force acting on the rod as shown. 1. Identify the pivot (left end of rod) 1.25m pivot 57o 2. Identify the force ( F = 16 N) 3. Determine the lever arm ( r = 1.25 m) F = 16N 4. Determine the angle between the force and the lever arm (q = 57o) 5. Calculate the torque: t = F sin(q )r = (16N )sin(57 )(1.25m) = 16.8N m o EXAMPLE: Calculate the torque produced by the force acting on the rod as shown. 1. Identify the pivot (left end of rod) 1.25m pivot 2. Identify the force ( F = 16 N) 3. Determine the lever arm ( r = 1.25 m) Rotation will be counter-clockwise F = 16N 4. Determine the angle between the force and the lever arm (q = 57o) 5. Calculate the torque: t = F sin(q )r = (16N )sin(57 )(1.25m) = 16.8N m o 6. Determine the orientation of the torque. (see diagram) 57o CENTER OF GRAVITY: The center of gravity of an object is the point, (xcg , ycg), from which the weight acts. xcg xmg = m g i i i i ycg i ymg = m g i i i i i For most ordinary objects all gi are equal. Thus we can write: x cg xm = m i i i y cg ym = m i i i When g is considered constant cg is usually referred to as center of mass. EXAMPLE: A system of three masses, m1, m2 and m3 are arranged as shown. Calculate the coordinates of the center of mass of this system. y m2 = 4 kg at (0, 3m) m3 = 3 kg at (4m, 0) x m1 = 5 kg at (0, 0) ANSWER: y A system of three masses, m1, m 2 and m3 are arranged as shown. Calculate the coordinates of the center of mass of this system. m2 = 4 kg at (0, 3m) m3 = 3 kg at (4m, 0) x m1 = 5 kg at (0, 0) xcg = m1 x1 m2 x2 m3 x3 m1 m2 m3 xcg = (5kg )(0m) (4kg )(0m) (3kg )(4m) (5kg )(0m) (4kg )(3m) (3kg )(0m) ycg = 5kg 4kg 3kg 5kg 4kg 3kg xcg = 1m m1 y1 m2 y2 m3 y3 ycg = m1 m2 m3 ycg = 1m MOMENT OF INERTIA: Objects have an intrinsic resistance to changes in their rate of rotation. This is referred to as rotational inertia and is quantified by moment of inertia, I. Moment of inertia depends on two factors: 1. the object’s mass, and 2. how that mass is distribute relative to the axis of rotation. I = mi ri 2 For many solid objects there are simple formulas for their moments of inertia. 2 2 Any axis through center. sphere 5 I = MR I disk = MR 1 2 2 Axis through center and perpendicular to disk. EXAMPLE: Suppose the three masses in the diagram are fixed together by massless rods and are free to rotate in the xy plane about their center of gravity. The coordinates of their center of gravity are (1m, 1m) as calculated in the previous example. Calculate the moment of inertia of these masses about this axis. y m2 = 4 kg at (0, 3m) r2 r1 r3 m3 = 3 kg at (4m, 0) x m1 = 5 kg at (0, 0) ANSWER: y Suppose the three masses in the diagram are fixed together by massless rods and are free to rotate in the xy plane about their center of gravity. The coordinates of their center of gravity are (1m, 1m) as calculated in the previous example. m2 = 4 kg at (0, 3m) r2 r1 r3 m3 = 3 kg at (4m, 0) x Calculate the moment of inertia of these masses about this axis. r1 = 1m 0m 1m 0m = 1.41m r2 = 1m 0m 1m 3m r3 = 1m 4m 1m 0m 2 2 2 m1 = 5 kg at (0, 0) 2 2 2 = 2.24m = 3.16m I = m1r12 m2 r22 m3r32 I = (5kg )(1.41m) 2 (4kg )(2.24m) 2 (3kg )(3.16m) I = 60.00kg m 2 2 ROTATIONAL DYNAMICS I: If the net torque acting on an object is not zero, the object will experience an angular acceleration, a. a is related to torque by Newton’s second law of motion for rotation: t = Ia EXAMPLE: An object has a moment of inertia of 60 kgm2 and is acted upon by a torque of 20Nm. What is the magnitude of the resulting angular acceleration of the object? ANSWER: An object has a moment of inertia of 60 kgm2 and is acted upon by a torque of 20Nm. What is the magnitude of the resulting angular acceleration of the object? t = Ia t 20 N m a= = 2 I 60kg m a = 0.33 rad s 2 ROTATIONAL DYNAMICS II: A rotating object has kinetic energy due to its rotational motion in addition to whatever kinetic energy it has due to the linear motion of its center of gravity. 2 1 rot 2 The work-energy theorem applies in the following way: KE = Iw DKElin DKErot DPE = WNC An object’s total kinetic energy is given by: KEtot = KElin KErot EXAMPLE: A solid ball rolls down a 40o incline from a height of 4m without slipping. The radius of the ball is 20cm and it begins from rest. What is the linear speed of the ball at the bottom of the incline? ANSWER: A solid ball rolls down a 40o incline from a height of 4m without slipping. The radius of the ball is 20cm and it begins from rest. What is the linear speed of the ball at the bottom of the incline? DKELin DKERot DPE = 0 KELin ,i = KERot ,i = 0 KELin , f = 12 mv 2 Basic Equation Values and expressions for initial and final quantities KERot , f = 12 Iw 2 2 1 2 v 2 1 2 = 2 5 mr 2 = 5 mv r PEi = mgh PE f = 0 Continued on next slide Continues from previous slide 1 2 mv mv mgh = 0 Substitutions made to produce working equation v 2 = gh Simplify and solve for v 7 10 2 1 5 2 10 gh v= = 7 v = 7.48 ms 10 9.8 sm2 4m 7 The speed of the ball does not depend on its mass nor on its radius. It even does not depend on the angle of the incline, just the height from which it starts. Do the same calculation with a cylinder and determine which will reach the bottom of the incline first, the sphere or the cylinder if released together. ANGULAR MOMENTUM: A rotating object has rotational momentum called angular momentum, L. L = Iw Angular momentum and linear momentum are different physical quantities and thus do not add. Whenever a system’s rotational motions are governed exclusively by internal torques, the system’s total angular momentum will be constant. Under such circumstances angular momentum is conserved. Li = Lf . EXAMPLE: A merry-go-round on a playground has a radius or 1.5 m and a mass of 225 kg. One child is sitting on its outer edge as it rotates 1 rev/s. If the mass of the child is 50 kg, what will the new rotation rate be when the child crawls half way to the center of the merry-go-round? ANSWER: A merry-go-round on a playground has a radius or 1.5 m and a mass of 225 kg. One child is sitting on its outer edge as it rotates 1 rev/s. If the mass of the child is 50 kg, what will the new rotation rate be when the child crawls half way to the center of the merry-go-round? L i = Lf I system,iw i = I system , f w f 1 2 2 mgr mgr m r wf = 1 2 1 2 m r 2 c c ,i w = i 2 mmgr rmgr mc rc2,i 2 mgr mgr m r 1 2 2 mgr mgr m r m r 2 c c, f w f wi m r 2 c c, f 225kg 1.5m 50kg 1.5m rad wf = 2 p s 2 2 1 2 225kg 1.5m 50kg .75m 2 1 2 w f = 8.17 rads f = 1.3Hz or 2 rev s EQUILIBRIUM: The forces acting on an object in equilibrium must satisfy both the first and second conditions of equilibrium. F x =0 F y =0 t = 0 All An object in equilibrium has no linear and no angular accelerations. That does not mean it isn’t moving, just not accelerating. When an object is stationary it is said to be in static equilibrium. Many of the equilibrium problems here will be static equilibrium problems. THE SECOND CONDITION OF EQUILIBRIUM: An object in rotational equilibrium has uniform angular velocity, w = constant. This means its angular acceleration, a, is zero. In order for a to be zero, the net or total torque acting on the object must be zero. t i =0 t i = ti = ti All All CCW CCW CW ti ti CW EXAMPLE: A uniform rod of length l = 4m and mass m = 75kg is hinged to a wall at the left and supported at the right by a cable. The cable is attached to the rod l/4 from its right edge. The rod makes a 30o angle with the horizontal and the cable makes a 45o angle with the rod. Calculate the tension in the cable and the force exerted on the rod by the hinge. cable wall rod 45o 30o hinge Solution on next several slides STEP 1: cable wall T rod 45o 30o R hinge W Add Force Vectors: T = Tension, W = Weight, R = Reaction STEP 1: 165O cable wall 210O T rod 45 pivot o 30o R hinge W Add Force Vectors: T = Tension, W = Weight, R = Reaction Add needed angles. STEP 1: 165O cable wall 210O T rod 45o 3l/4= 3m 30o R hinge l/2 = 2m W Add Force Vectors: T = Tension, W = Weight, R = Reaction Add needed angles. Add Lever Arms. STEP 2: Force analysis F cos(165o )iˆ F sin(165 o ) ˆj W= 0iˆ m gjˆ R= Rxiˆ R y ˆj F= First Condition of Equilibrium: F = Rx T cos165 = 0 F = Ry 735 N T sin165 = 0 x y o o On the next slide these force equations will be used to solve for Rx and Ry. STEP 3: Torque Analysis To analyze the torques acting on the rod, construct to following table or small spreadsheet. The column headings are: F = force F = force component perpendicular to lever arm r = lever arm t = the torque, which is F r dir = direction of torque (CW or CCW) The entries in each row correspond to values associated with one force on the rod. One row for each force. The next slide continues the example solution. F F r t dir g = 9.8 m/s2 was used STEP 3: Torque Analysis F F r T T sin(45o ) W mg sin(60o ) R 3l 4 l 2 t dir 3Tl sin(45o ) CCW 4 mgl sin(60o ) CW 2 0 Second Condition of Equilibrium: t CCW = t CW T 2.1213m = 1273.06 Nm T = 600.13N ANSWER From force equations on previous slide and using T=600.13N: Rx=579.68N Ry=579.67N Then using the Pythagorean Theorem and tan-1: R = 819.8N q = 45o ANSWER