Rotational Motion. Rotational Motion. Rotational Motion
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1 I. Rotational Kinematics II. Rotational Dynamics (Netwton s Law for Rotation) III. Angular Momentum Conservation 1. Remember how Newton s Laws for translational motion were studied: 1. Kinematics (x = x 0 + v 0 t + ½ a t ). Dynamics (F = m a) 3. Momentum Conservation. Now, we repeat them again, but for rotational motion: 1. Kinematics (θ, ω, α). Dynamics (τ = I α) 3. Angular Momentum Part I Translational Motion and Today Later 1
2 Newton s s Laws for Rotation τ = Iα net nd part 1 st part 3rd part Angular Quantities Kinematical variables to describe the rotational motion: Angular position, velocity and acceleration l θ = ( rad) R θ dθ ω = lim = t 0 t dt ω ave ω dω α = lim = t 0 t dt α ave ( rad/s) ( rad/s )
3 Linear and Angular Quantities a tan a rad Vector Nature of Angular Quantities Kinematical variables to describe the rotational motion: Angular position, velocity and acceleration c.w. or c.c.w. rotation (like +x or x direction in 1D) Vector natures! l θ = R d kˆ θ ω = kˆ dt d kˆ ω α = kˆ dt ( rad) ( rad/s) ( rad/s >0 or <0 ) x z R.-H. Rule y 3
4 Example 1 ( (α = constant) (1) What is the angular speed of rotation of the Earth? 1 rotation (π rad) per 4 hours (1440 min) () An old 33 1/3 rpm record player starts from rest and reaches operating speed in.00 seconds. Through what angle did it turn in those.00 seconds? t =.00 s, ω 0 = 0, ω = 33 1/3 rpm (3) A computer hard drive rotates at 5400 rpm. What angular acceleration will get it up to speed in just 150 revolutions starting from rest? Use Eq. (3) with θ θ 0 = 150 rev. =? rad Example 1 (cont d) (4) A hard drive reaches 5400 rpm in 3.0 seconds. What was the average angular speed assuming constant angular acceleration? (5) A dentist s drill accelerates to 1800 rpm in.50 seconds. what is its angular acceleration? (6) The angular velocity changes from 47.0 rad/s to 47.0 rad/s in.00 seconds. What is the angular acceleration? 4
5 Example A grinding wheel turns at a constant angular acceleration of 60.0 rad/s from 4.0 rad/s for.00 sec. Then, a circuit breaker trips. It turns through 43 rad as it coasts to a stop at a constant angular acceleration. Find: (a) the total angle between t = 0 and the time it stopped; (b) the time it stopped; (c) the angular acceleration as it slowed down. Also sketch θ-t graph. Practice Problem 1 Problem 3: (5 points) A grinding wheel turns at a varying angular acceleration of α(t) = [30.0 rad/s 3 ] t for.00 sec. Assume the initial angular speed of 0.0 rad/s. Then, a circuit breaker trips. It turns through 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. 5
6 Practice Problem A solid cylinder (radius R = m and height H = 5 m) turns at a constant angular acceleration of 60.0 rad/s from 4.0 rad/s for.00 sec. Then, a circuit breaker trips. It turns through 43 rad as it coasts to a stop at a constant angular acceleration. (a) Find the total angle between t = 0 and the time it stopped. (b) Find the time it stopped. (c) Find the angular acceleration as it slowed down. 0.4H (d) Find the speed (v) of point P at t =.00 sec. (e) Sketch the motion of point P in v-t graph. Also sketch θ-t graph. ω H 1. Remember how Newton s Laws for translational motion were studied: 1. Kinematics (x = x 0 + v 0 t + ½ a t ). Dynamics (F = m a) 3. Momentum Conservation. Now, we repeat them again, but for rotational motion: 1. Kinematics (θ, ω, α). Dynamics (τ = I α) 3. Angular Momentum Part II Translational Motion and Today Later 6
7 Torque due to Gravity? τ = r F =? We assume: Center of mass (CM or cm) = Center of gravity (if uniform gravity) Torque due to Gravity? τ = r F =? mass m 7
8 Net Torque due to Gravity? One extra sphere! Do we need a new concept? No! mass m l τ net = τ rod + τ sphere Solid sphere (M, r) τ τ = F net ( R Note: sign of τ and Στ τ 1 = F 1 ( R 1 sin90 ) = (50.0N)(0.300 m) = 15.0N m = (50.0N)(0.500 m)(0.866) = 1.7N m = τ [c.c.w.] + τ [c.w.] 1 = τ [ + 1] + τ [ 1] 1 sin60 ) = [(15.0N m) (1.7N m)][c.c.w.] = 6.7N m[c.c.w.] 6.7N m[c.w.] r 1 r 8
9 Example 1 Calculate the torque on the.00-m long beam due to a 50.0 N force (top) about (a) point C (= c.m.) (b) point P Calculate the torque on the.00-m long beam due to a 60.0 N force about (a) point C (= c.m.) (b) point P Calculate the torque on the.00-m long beam due to a 50.0 N force (bottom) about (a) point C (= c.m.) (b) point P Example 1 (cont d) Calculate the net torque on the.00-m long beam about (a) point C (= c.m.) (b) point P 9
10 Ptactice Problem Determine the net torque (magnitude and direction) on the.00-m-long beam about: 50.0 N a. (15 pts) point C (the CM position); θ = 70.0 b. (10 pts) point P at one end. o τ i = ri Fi =? τ τ net = i i 60.0 N θ = 50.0 o C P θ = 33.0 o 50.0 N Ptactice Problem Determine the net torque (magnitude and direction) on the disk about pin. τ i = ri Fi =? τ τ net = i i F 1 = 50N θ 1 =40 o θ =70 o F = 5N 10
11 Newton s s Laws for Rotation τ = Iα net nd part [N m] 3rd part 1 st part [s ] Parallel-axis axis Theorem d d 1 I = I 1 + I 11
12 (a) Express the moment of inertia of the array of point objects about the y-axis in terms of m, M, X 1, X, and/or Y. Y X 1 X Newton s s Laws for Rotation τ = Iα net nd part [N m] 3 rd part [kg m ] 1 st part [s ] K rot = (1/) I ω ( K = (1/) m v ) Application? 1
13 Rolling Motion (w/o slipping) PHYSICS, R. Serway and R. Beichner, 5 th ed. Rolling Motion w/o Slipping () ω ω Faster Instantaneously rest Instantaneous axis 13
14 Example 3 What will be the speed of each of the following objects when it reaches the bottom of an incline if it starts from rest at a vertical height H and rolls without slipping? Find the speed using 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) Example 4 Analyze the rolling sphere in terms of forces and torques: find the magnitudes of the velocity v and the friction force F fr. 1. F.B.D.? Rolling Motion (w/o slipping). Is F fr static or kinetic friction? Static Friction No work No energy loss K+U=constant P is: (a) the point of contact with the ground; (b) instantaneously rest relative to the ground. Thus: The motion of the wheel is a pure rotation about the instantaneous axis through P at the instant. 14
15 Example 5 A sting is wrapped around a uniform solid cylinder of mass M (= kg) and radius R (= 0.00 m), and the cylinder starts falling from rest. Draw the freebody diagram for the cylinder while it descends. A l s o f i n d ( a ) i t s acceleration, (b) the tension in the string, and (c) the speed after the cylinder has descended h = m. Example 5 - Challenge A sting is wrapped around a hollow cylinder of mass M (= kg), inner radius R 1 (= m), outer radius R (= 0.00 m). The cylinder starts falling from rest. (a) Draw the free-body diagram for the cylinder while it descends. Also find: (b) its acceleration; (c) the tension in the string; (d) the speed after the cylinder has descended h = m. 15
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