Rotational Motion

Rotational MotionI. Rotational KinematicsII. Rotational Dynamics(Netwton’s Law for Rotation)III. Angular Momentum ConservationRotational Motion1. Remember how Newton’s Laws for translationalmotion were studied:1. Kinematics (x = x 0 + v 0 t + ½ a t 2 )2. Dynamics (F = m a)3. Momentum Conservation2. Now, we repeat themagain, but for rotationalmotion:1. Kinematics (θ, ω, α)2. Dynamics (τ = I α)3. Angular MomentumPart IRotational MotionTranslational Motion andRotational MotionToday Later1

Newton’s s Laws for Rotation τ = Iαnet2 nd part1 st part3rd partRotational MotionAngular QuantitiesKinematical variables to describe the rotational motion:Angular position, velocity and accelerationlθ = ( rad)R∆θdθω = lim =∆t→0∆t dtωave∆ωdωα = lim =∆t→0∆t dtαave( rad/s)( rad/s2)Rotational Motion2

Linear and Angular Quantitiesa tana radRotational MotionVector Nature ofAngular QuantitiesKinematical variables to describe the rotational motion:Angular position, velocity and accelerationc.w. or c.c.w. rotation (like +x or –x direction in 1D)Vector natures!lθ =Rdkˆθω = kˆdtdkˆωα = kˆdt( rad)( rad/s)( rad/s>0 or

Example 1 ( (α = constant)(1) What is the angular speed of rotation of the Earth?1 rotation (2π rad) per 24 hours (1440 min)(2) An old 33 1/3 rpm record player starts from rest andreaches operating speed in 2.00 seconds. Through whatangle did it turn in those 2.00 seconds?t = 2.00 s, ω 0 = 0, ω = 33 1/3 rpm(3) A computer hard drive rotates at 5400 rpm. Whatangular acceleration will get it up to speed in just 150revolutions starting from rest?Use Eq. (3) with θ – θ 0 = 150 rev. = ? radRotational MotionExample 1 (cont’d)(4) A hard drive reaches 5400 rpm in 3.20 seconds. Whatwas the average angular speed assuming constantangular acceleration?(5) A dentist’s drill accelerates to 1800 rpm in 2.50 seconds.what is its angular acceleration?(6) The angular velocity changes from 47.0 rad/s to –47.0rad/s in 2.00 seconds. What is the angular acceleration?Rotational Motion4

Example 2A grinding wheel turns at a constant angular accelerationof 60.0 rad/s 2 from 24.0 rad/s for 2.00 sec. Then, acircuit breaker trips. It turns through 432 rad as it coaststo a stop at a constant angular acceleration.Find:(a) the total angle between t = 0and the time it stopped;(b) the time it stopped;(c) the angular acceleration as itslowed down.Also sketch θ-t graph.Rotational MotionPractice Problem 1Problem 3: (25 points)A grinding wheel turns at a varying angular acceleration of α(t) = [30.0 rad/s 3 ] t for 2.00sec. Assume the initial angular speed of 20.0 rad/s. Then, a circuit breaker trips. It turnsthrough 400 rad as it coasts to a stop at a constant angular acceleration.a. (5 pts) Find the total angle (θ total ) between t = 0 and the time it stopped.b. (10 pts) Find the time (t total ) it stopped. Find the angular acceleration as it slowed down.c. (10 pts) Sketch θ−t, ω-t, α-t graphs.Rotational Motion5

Practice Problem 2A solid cylinder (radius R = 2 m and height H = 5 m) turns at aconstant angular acceleration of 60.0 rad/s 2 from 24.0 rad/sfor 2.00 sec. Then, a circuit breaker trips. It turns through432 rad as it coasts to a stop at a constant angular acceleration.(a) Find the total angle between t = 0and the time it stopped.(b) Find the time it stopped.(c) Find the angular accelerationas it slowed down.0.4H(d) Find the speed (v) of point Pat t = 2.00 sec.(e) Sketch the motion of point Pin v-t graph. Also sketch θ-t graph.ωHRotational Motion1. Remember how Newton’s Laws for translationalmotion were studied:1. Kinematics (x = x 0 + v 0 t + ½ a t 2 )2. Dynamics (F = m a)3. Momentum Conservation2. Now, we repeat themagain, but for rotationalmotion:1. Kinematics (θ, ω, α)2. Dynamics (τ = I α)3. Angular MomentumPart IIRotational MotionTranslational Motion andRotational MotionToday Later6

Torque due to Gravity? τ = r × F= ?We assume:Center of mass (CM or cm)= Center of gravity(if uniform gravity)Rotational MotionTorque due to Gravity? τ = r × F= ?mass mRotational Motion7

Net Torque due to Gravity?One extra sphere!Do we need a new concept?No!mass mlτnet= τrod+ τsphereSolid sphere (M, r)Rotational Motionττ = Fnet2( R 22⋅Note: sign of τ and Σττ 1= F 1( R 1sin90 )= (50.0N)(0.300 m) = 15.0N⋅m= (50.0N)(0.500 m)(0.866)= 21.7N m = τ [c.c.w.]+ τ [c.w.]1= τ [ + 1] + τ [ −1]1sin60 )22= [(15.0N⋅m)− (21.7N⋅m)][c.c.w.]= − 6.7N⋅m[c.c.w.]→6.7N⋅m[c.w.]Rotational Motionr 1r 28

Example 1Calculate the torque on the 2.00-m longbeam due to a 50.0 N force (top) about(a) point C (= c.m.)(b) point PCalculate the torque on the 2.00-m longbeam due to a 60.0 N force about(a) point C (= c.m.)(b) point PCalculate the torque on the 2.00-m longbeam due to a 50.0 N force (bottom) about(a) point C (= c.m.)(b) point PRotational MotionExample 1 (cont’d)Calculate the net torque on the 2.00-mlong beam about(a) point C (= c.m.)(b) point PRotational Motion9

Ptactice Problem 2Determine the net torque (magnitude and direction) on the2.00-m-long beam about:50.0 Na. (15 pts) point C (the CM position);θ = 70.0b. (10 pts) point P at one end.o τi= ri× Fi= ? τ τnet=ii60.0 Nθ = 50.0 oCPθ = 33.0 oRotational Motion50.0 NPtactice Problem 2Determine the net torque (magnitude and direction) on thedisk about pin. τi= ri× Fi= ? τ τnet=iiF 1= 50Nθ 1 =40 oθ 2 =70 oF 2= 25NRotational Motion10

Newton’s s Laws for Rotation τ = Iαnet2 nd part[N m]3rd part1 st part[s –2 ]Rotational MotionParallel-axis axis Theoremd 22d 1I = I 1+ IRotational Motion11

2(a) Express the moment of inertia of the array of pointobjects about the y-axis in terms of m, M, X 1 , X 2 , and/orY.X 1 X 2YRotational MotionNewton’s s Laws for Rotation τ = Iαnet2 nd part[N m]3 rd part[kg m 2 ]1 st part[s –2 ]K rot = (1/2) I ω 2 ( K = (1/2) m v 2 )Application?Rotational Motion12

Rolling Motion (w/o slipping)“PHYSICS,” R. Serway and R. Beichner, 5 th ed.Rotational MotionRolling Motion w/o Slipping (2)ωωFasterRotational MotionInstantaneously restInstantaneous axis13

Example 3What will be the speed of each of the following objects whenit reaches the bottom of an incline if it starts from rest at avertical height H and rolls without slipping? Find the speedusing the conservation of mechanical energy.(i) Solid sphere (M, R 0 )(ii) Solid cylinder (M, R 0 , l)(iii) Thin hoop (M, R 0 , l)Rotational MotionExample 4Analyze the rolling sphere in terms of forces and torques: findthe magnitudes of the velocity v and the friction force F fr .1. F.B.D.?Rolling Motion(w/o slipping)2. Is F fr static or kinetic friction?Static FrictionNo workNo energy lossK+U=constantRotational MotionP is:(a) the point of contact with theground;(b) instantaneously rest relativeto the ground.Thus:The motion of the wheel is a purerotation about the instantaneousaxis through P at the instant.Rotational Motion14

Example 5A sting is wrapped arounda uniform solid cylinder ofmass M (= 0.500 kg) andradius R (= 0.200 m), andthe cylinder starts fallingfrom rest. Draw the freebodydiagram for thecylinder while it descends.A l s o f i n d ( a ) i t sacceleration, (b) the tensionin the string, and (c) thespeed after the cylinder hasdescended h = 0.800 m.Rotational MotionExample 5 - ChallengeA sting is wrapped around a hollow cylinder of mass M(= 0.500 kg), inner radius R 1 (= 0.100 m), outer radiusR 2 (= 0.200 m). The cylinder starts falling from rest.(a) Draw the free-body diagram for the cylinder while itdescends.Also find:(b) its acceleration;(c) the tension in the string;(d) the speed after the cylinder hasdescended h = 0.800 m.Rotational Motion15