This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: lee (cyl327) – HW10 – gentle – (56245) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A bowling ball is given an initial speed v on an alley such that it initially slides without rolling. The coefficient of friction between ball and alley is μ . Find the speed of the ball’s center of mass v CM at the time pure rolling motion occurs. 1. v CM = 5 7 v correct 2. v CM = 3 7 v 3. v CM = 3 5 v 4. v CM = 1 3 v 5. v CM = 1 5 v Explanation: Basic Concepts: F fr = μN L = I ω . Solution: The frictional force F = μM g acts to slow the ball v = v F M t. (1) It also exerts a torque RF and starts the ball rolling I ω = RF t. The ball starts rolling without slipping at time t = τ , when ω = v R ω = v R = v F M τ R = RF I τ . Solve for τ with I = 2 5 M R 2 and F = μM g τ = v R RF I + F M R = 2 7 v μg . Substituting the expression for τ into (1) yields v ( τ ) = v μM g M τ = 5 7 v . 002 (part 2 of 2) 10.0 points Since the frictional force provides the decel eration, from Newton’s second law it follows that a CM = μg . Find the distance x it has traveled. When pure rolling motion occurs, v CM = Rω . 1. x = 2 v 2 7 μg 2. x = 12 v 2 49 μg correct 3. x = 12 v 2 25 μg 4. x = 3 v 2 14 μg 5. x = 5 v 2 81 μg Explanation: The distance traveled by the ball from t = 0 to t = τ is x = integraldisplay t = τ t =0 v dt = integraldisplay t = τ t =0 ( v μg t ) dt = 12 49 v 2 μg keywords: 003 (part 1 of 2) 10.0 points A string is wound around a uniform disc of radius 0 . 53 m and mass 3 kg . The disc is released from rest with the string vertical and its top end tied to a fixed support. The acceleration of gravity is 9 . 8 m / s 2 . lee (cyl327) – HW10 – gentle – (56245) 2 h . 53 m 3 kg ω As the disc descends, calculate the tension in the string. Correct answer: 9 . 8 N. Explanation: Let : R = 0 . 53 m , M = 3 kg , and g = 9 . 8 m / s 2 . Basic Concepts summationdisplay vector F = mvectora summationdisplay vector τ = I vectorα Δ U + Δ K rot + Δ K trans = 0 Solution summationdisplay F = T M g = M a and (1) summationdisplay τ = T R = I α = 1 2 M R 2 parenleftBig a R parenrightBig . (2) Solving for a in (2), a = 2 T M . (3) Using a from Eq. (3) and solving for T in (1), T = M ( g a ) = M parenleftbigg g 2 T M parenrightbigg = M g 2 T 3 T = M g T = M g 3 (4) = (3 kg) (9 . 8 m / s 2 ) 3 = 9 . 8 N . 004 (part 2 of 2) 10.0 points Calculate the speed of the center of mass when, after starting from rest, the center of mass has fallen 3 . 5 m. Correct answer: 6 . 76264 m / s. Explanation: From conservation of mechanical energy we have Δ U + Δ K rot + Δ K trans = 0 1 2 parenleftbigg 1 2 M R 2 parenrightbigg ω 2 + 1 2 M v 2 = M g Δ h....
View
Full
Document
This note was uploaded on 02/28/2012 for the course PHY 316 taught by Professor Markert during the Fall '10 term at University of Texas.
 Fall '10
 MARKERT
 mechanics

Click to edit the document details