HS000011

HS000011 - Busto Anthony – Homework 11 – Due 11:00 pm...

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Unformatted text preview: Busto, Anthony – Homework 11 – Due: Nov 24 2006, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A constant horizontal force of 200 N is applied to a lawn roller in the form of a uniform solid cylinder of radius 0 . 31 m and mass 12 kg. The acceleration of gravity is 9 . 8 m / s 2 . M R F If the roller rolls without slipping, calculate the acceleration of the center of mass. ( I cm = 1 2 M R 2 ) Correct answer: 11 . 1111 m / s 2 . Explanation: Basic Concepts: X ~ F = m~ a X ~ τ = I~α If we choose the center of mass as the axis, X F = F- f = M a τ = fR = I cm α = 1 2 M R 2 a R then you can solve these two equations to get a. The another way to do this is we can choose the contact point as the axis, using I = 1 2 M R 2 + M R 2 = 3 2 M R 2 , we find 3 2 M R 2 a R = F R . Therefore a = 2 F 3 M (1) = 2(200 N) 3(12 kg) = 11 . 1111 m / s 2 . keywords: 002 (part 1 of 1) 10 points A 200 g basketball has a 14 . 6 cm diameter and may be approximated as a thin spherical shell. Given: The moment of inertia of a thin spherical shell of radius R and mass m is I = 2 3 m R 2 , g = 9 . 8 m / s 2 , and μ = 0 . 44 . 200 g μ = . 4 4 1 4 . 6 c m 2 . 1 5 m 20 ◦ Starting from rest, how long will it take a basketball to roll, without slipping, 2 . 15 m down an incline that makes an angle of 20 ◦ with the horizontal? Correct answer: 1 . 46224 s. Explanation: Given : ‘ = 2 . 15 m , θ = 20 ◦ , and R = d 2 = 14 . 6 cm 2 = 0 . 073 m . P m g s i n θ N m g θ The moment of inertia of the ball about the point of contact between the ball and the inclined plane is I P = I cm + m d 2 = 2 3 m R 2 + m R 2 = 5 3 m R 2 . Busto, Anthony – Homework 11 – Due: Nov 24 2006, 11:00 pm – Inst: Opyrchal, H 2 The net torque about the point of contact between the ball and the inclined plane is m g R sin θ = I P α = 5 3 m R 2 α so a = m g R 2 sin θ 5 3 m R 2 = 3 5 g sin θ . If the sphere starts from rest, then its center of mass moves a distance ‘ = 1 2 a t 2 , which gives t = r 2 ‘ a = s 2(2 . 15 m) (2 . 01108 m / s 2 ) = 1 . 46224 s ....
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This note was uploaded on 04/03/2009 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.

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HS000011 - Busto Anthony – Homework 11 – Due 11:00 pm...

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