# PHYSICS 111 HOMEWORK SOLUTION #9

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PHYSICS 111 HOMEWORKSOLUTION #9April 5, 2013

0.1A potterŌĆÖs wheel moves uniformly from rest to an angular speed of0.16 rev/s in 33 s. a) Find its angular acceleration in radians per second per second. b) Would doubling the angular acceleration during the givenperiod have doubled final angular speed?a)Žē Žē0 ╬▒tFrom rest to an angular speed of 0.16 rev/s in 33s we should have:╬▒ ŽētŽē Žē0 t0.16 2╬Ā 33 0.030 rad/s2 b)For double the angular acceleration we should have :2╬▒ 2 Žē2 Žē ttThe angular speed will be doubled as well2

0.2.0.2During a certain time interval, the angular position of a swingingdoor is described by ╬Ė 4.95 9.4t 2.05t2 , where ╬Ė is in radiansand t is in seconds. Determine the angular position, angular speed,and angular acceleration of the door at the following times. a) t 0 b) t 2.99 sa) at t 01 ╬Ė0 Žē0 t ╬▒t22 4.95 9.4t 2.05t2╬Ė╬Ė 4.95 radŽē Žē0 ╬▒t 9.4 4.1t 9.4 rad/s╬▒ 4.1 rad/s2b) at t 2.99 s╬Ė╬Ė1 ╬Ė0 Žē0 t ╬▒t22 4.95 9.4t 2.05t2 4.95 9.4 2.99 2.05 2.992 51.38 radŽē Žē0 ╬▒t 9.4 4.1t 9.4 4.1 2.99 21.66 rad/s╬▒ 4.1 rad/s23

0.3An electric motor rotating a workshop grinding wheel at 1.06 102rev/min is switched off. Assume the wheel has a constant negativeangular acceleration of magnitude 1.96 rad/s2 . a)How long does it take the grinding wheel to stop? b) Through how many radians has the wheel turned during thetime interval found in part (a)?a)The motor having an initial angular speed of Žē0 1.06 102 rev/min willcome to a stop when its angular speed goes down to zero:Žē Žē0 ╬▒t0 t2ŽĆ 1.96t600 1.06 102 2ŽĆ60 1.96 5.66 s1.06 102 b)The radians covered between switching off and stopping is ╬Ė ╬Ė ╬Ė0╬Ė ╬Ė1 ╬Ė0 Žē0 t ╬▒t222ŽĆ1.96 22 1.06 10 t t6022ŽĆ1.96 1.06 102 5.66 5.662602 431.43 rad

0.4.0.4A racing car travels on a circular track of radius 275 m. Suppose thecar moves with a constant linear speed of 51.5 m/s. a)Find its angular speed. b)Find the magnitude and direction of its acceleration.a)Angular and linear speed are always related through : v rŽēŽēvr51.5 275 0.19 rad/s With a constant linear speed the acceleration is radial (a ar dvdt 0) :a v2ras at v2r51.522759.645 m/s25

0.5A discus thrower accelerates a discus from rest to a speed of 25.3 m/sby whirling it through 1.28 rev. Assume the discus moves on the arcof a circle 0.99 m in radius. a) Calculate the final angular speed of the discus. b) Determine the magnitude of the angular acceleration of thediscus, assuming it to be constant. c) Calculate the time interval required for the discus to accelerate from rest to 25.3 m/s.a)The final angular speed is again related to the final linear speed as :vŽē r25.3 0.99 25.56 rad/sb)From rest to 25.56 rad/s, Angular acceleration assumed to be constant can beevaluated from the time-independent equation:Žē 2 Žē02╬▒62╬▒ ╬ĖŽē 2 Žē02 2 ╬Ė25.562 0 2 1.28 2ŽĆ 40.60 rad/s2

0.6.c)1 ╬Ė0 Žē0 t ╬▒t221 2 ╬Ė ╬▒t2r2 ╬Ėt ╬▒r2 1.28 2ŽĆ 40.60 0.63 s╬Ė0.6The figure below shows the drive train of a bicycle that has wheels67.3 cm in diameter and pedal cranks 17.5 cm long. The cyclist pedalsat a steady cadence of 79.0 rev/min. The chain engages with a frontsprocket 15.2 cm in diameter and a rear sprocket 6.00 cm in diameter. a) Calculate the speed of a link of the chain relative to thebicycle frame. b) Calculate the angular speed of the bicycle wheels. c) Calculate the speed of the bicycle relative to the road. d) What piece of data, if any, are not necessary for the calculations?7

a)Links of the chain rotate around the front sprocket. The front sprocket rotates at 79.0 rev/min and has a diameter of 15.2 cm. Links on the chain willconsequently have a linear speed of:v rf r Žēf r0.1502 79 2ŽĆ 260 0.621 m/sb)The bicycle wheels and the rear sprocket have the same angular speed that wecan calculate from the linear speed of the chain links (links rotate around therear sprocket also):Žērear vrrear0.6210.0620.7 rad/sc)Again the rear sprocket and the rear wheel rotating with the same angularspeed will give rise to a linear speed for both wheels and the bicycle:v rwheel Žērear0.673 20.72 6.97 m/sd)The only piece of data not necessary for our calculations is the length of pedalcranks.8

0.7.0.7A wheel 1.65 m in diameter lies in a vertical plane and rotates aboutits central axis with a constant angular acceleration of 3.70 rad/s2 .The wheel starts at rest at t 0, and the radius vector of a certainpoint P on the rim makes an angle of 57.3 with the horizontal atthis time. At t 2.00 s, find the following: a) the angular speed of the wheel. b) the tangential speed of the point P. c) the total acceleration of the point P. d) the angular position of the point P.a)The wheel started at rest. Therefore, Žē0 0:Žē Žē0 ╬▒t ╬▒t 3.70 2 7.40 rad/sb)The tangential speed of point P located on the rim:v rŽē1.65 7.402 6.11 m/sc)To calculate the total acceleration of the point P, we need to calculate boththe radial and tangential compenents:at dŽē1.65dv r r╬▒ 3.70 3.05 m/s2dtdt2ar v26.112 1.65 45.18 m/s2r29

and finally,a qa2t a2n45.28 m/s2Its direction ╬▓ with respect to the radius to P can be evaluated fromtan ╬▓ at3.05 an45.18i.e-╬▓ 3.86 d)╬Ė101╬Ė0 Žē0 t ╬▒t22ŽĆ1 57.3 0 3.70 221802 8.40 rad

0.8.0.8Rigid rods of negligible mass lying along the y axis connect threeparticles. The system rotates about the x axis with an angular speedof 2.10 rad/s. a) Find the moment of inertia about the x axis. b) Find the total rotational kinetic energy evaluated from 21 IŽē 2 . c) Find the tangential speed of each particle. d) Find the total kinetic energy evaluated fromP122 mi via)I X 4 32 2 22 3 42 92 kg.m2mi ri2b) 1 2IŽē20.5 92 2.102 202.86 JKi 11

c)Different linear speeds for different radius.However, all particles are rotatingat same angular speed: vi ri ŽēMass 1:v1 r1 Žē 3 2.10 6.30 m/sMass 2:v2 r2 Žē 2 2.10 4.20 m/sMass 3:v3 r3 Žē 4 2.10 8.40 m/sd)The total kinetic energy is:Ki X12mi vi21(4 6.302 2 4.202 3 8.402 )2202.86 Je)Both expressions lead to the same value.This equality can be easily proven:Ki 12X121Xmi vi2mi ri2 Žē 221 2XŽēmi ri221 2IŽē2

0.9.0.9A war-wolf or trebuchet is a device used during the Middle Ages tothrow rocks at castles and now sometimes used to fling large vegetables and pianos as a sport. A simple trebuchet is shown in the figurebelow. Model it as a stiff rod of negligible mass, d 3.20 m long,joining particles of mass m1 0.130 kg and m2 58.0 kg at its ends.It can turn on a frictionless, horizontal axle perpendicular to the rodand 15.0 cm from the large-mass particle. The operator releases thetrebuchet from rest in a horizontal orientation. a) Find the maximum speed that the small-mass object attains. b) While the small-mass object is gaining speed, does it movewith constant acceleration? c) Does it move with constant tangential acceleration? d) Does the trebuchet move with constant angular acceleration? e) Does it have constant momentum? f) Does the trebuchetEarth system have constant mechanicalenergy?a)LetŌĆÖs adopt the following notations: r1 and r2 the distance from the rotation axis to mass m1 and m2 respectively. we have r1 d r2 3.20 0.15 3.05m v designates the speed of m113

The frictionless motion of the system guarantees energy conservation: at restthis energy is zero. Once the heavier mass m2 is released to a lower positionfrom rest, mass m1 will rise and start gaining speed and higher gravitationalpotential energy. A maximum speed is reached when it is on a vertical position.Both masses will rotate with same angular speed Žē Energy conservation requires:0 1m1 gr1 m2 gr2 IŽē 22vmax 21)m1 gr1 m2 gr2 (m1 r12 mr22 )(2r1Rearranging should give :s2g(m2 r2 m1 r1 )vmax r1m1 r12 m2 r22r2 9.81(58 0.15 0.130 3.05) 3.050.130 3.052 58 0.152 24.55 m/sb)The overall acceleration is changing direction throughout the motion. Vectoracceleration is thus not constant.c) and d)Neither tangential acceleration nor angular acceleration are constant. First ofall, the two are related through at r1 ╬▒. LetŌĆÖs look at ╬▒ with second law :XŽänet I╬▒The net torque Žä must be changing from a maximum when the arm of thetrebuchet is horizontal and going to zero as it stands vertical. Remember thetorque expression : Žä F orce d, where d is the perpendicular distance fromthe axis of rotation to the vector force line and is not to be confused with r !!.Distance d is going from r1 to 0, changing the torque, the angular accelerationand the tangential acceleration.e)P We started our problem with an isolated system for whichFext 0. Thetrebuchet-earth system should preserve its mechanical energy.14

0.10.http://history.knoji.com/the-trebuchet/0.10A uniform, thin, solid door has height 2.00 m, width 0.825 m, andmass 23.5 kg. a) Find its moment of inertia for rotation on its hinges. b) Is any piece of data unnecessary?a)The moment of inertia of the thin door (considered a rectangular plate withwidth w and height h) rotating about the hinges axis is given by :I1mw2323.5 0.8252 3 5.33 kg.m2 b)The height of the door is not necessary for the calculation.15

0.11Many machines employ cams for various purposes, such as openingand closing valves. In the figure below, the cam is a circular disk ofradius R with a hole of diameter R cut through it. As shown in thefigure, the hole does not pass through the center of the disk. The camwith the hole cut out has mass M. The cam is mounted on a uniform,solid, cylindrical shaft of diameter R and also of mass M. What is thekinetic energy of the camshaft combination when it is rotating withangular speed Žē about the shaftŌĆÖs axis? (Use any variable or symbolstated above as necessary.)a)Our objects are considered uniform and the mass is distributed evenly. If thecam with the hole cut out has a mass M, we can easily convince ourselves thatthat the cam without the cut should have a mass of 43 M and the mass of thecut is 13 M . Moment of inertia of the cam without the hole cut out (a disk of mass Mand radius R)about the center is:1 42( M R2 ) M R22 33 Using the parallel axis theorem, the moment of inertia about the rotatingaxis is:24RM R 2 M ( )2 M R 2332 the cut-out hole (a disk of Mass 31 M and radiusabout the rotating axis :1M R 21( ) M R22 3 22416R2)momentŌĆÖs of inertia

0.12. The shaft (cylinder of mass M and radius R) momentŌĆÖs of inertia aboutthe rotating axis:1R1M ( )2 M R2228 Finally, the overall moment of inertia of our system about the rotatingaxis is :1113I M R2 (1 ) M R224 812The kinetic energy is :1 213IŽē M R2 Žē 22240.12Three identical thin rods, each of length L and mass m, are weldedperpendicular to one another as shown in the figure below. The assembly is rotated about an axis that passes through the end of onerod and is parallel to another. Determine the moment of inertia ofthis structure about this axis. (Use any variable or symbol statedabove as necessary.)a)Rotation axis is parallel to the y-axis with an offset distance of L2 . On the otherhand, the y-axis is a symmetry axis for the whole system of the three rods.Therefore, a good strategy is to calculate the moment of inertia of the threerods system about the y-axis and use the parallel axis theorem to determinethe moment of inertia about the given axis.17

The x-rod contributesdoes the z-rod.mL212as moment of inertia about the y-axis and so The y-rod, a thin cylinder, contributes zero about its own y-axis A total ofmL26The Moment of inertia of the system about the axis in consideration istherefore:I 18LI0 (3m)( )223mlmL2 64112mL12

max r 1 s 2g(m 2r 2 m 1r 1) m 1r2 1 m 2r22 3:05 r 2 9:81(58 0:15 0:130 3:05) 0:130 3:052 58 0:152 24:55 m s b) The overall acceleration is changing direction throughout the motion. Vector acceleration is thus not constant. c) and d) Neither tangential acceleration nor angular acceleration are constant. First of all, the two are .