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Kaplunovsky - hw 11

# Kaplunovsky - hw 11 - homework 11 Due 4:00 am Question 1...

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homework 11 – – Due: Apr 12 2007, 4:00 am 1 Question 1, chap 13, sect 2. part 1 of 2 8 points A spool of wire of mass M = 5 . 4 kg and radius R = 0 . 64 m is unwound under a constant wire tension F = 4 . 3 N. Assume the spool is a uniform solid cylinder that rolls without slipping. R F M Calculate the friction force f on the bottom of the spool. Take the rightward direction to be positive. Correct answer: 1 . 43333 N (tolerance ± 1 %). Explanation: By Newton’s Second Law, the spool’s cen- ter of mass (which is in the middle of the spool’s axis) accelerates according to Ma = F + f (1) where F is the wire tension, f is the friction force at the bottom of the spool, and the pos- itive direction for both forces is towards the right. At the same time, the spool’s angu- lar acceleration obtains from the net torque in the center-of-mass frame, thus = F × R - f × R (2) where I I cm = MR 2 2 . (3) Finally, since the spool rolls without slipping, α = a R . (4) Substituting eqs. (3) and (4) into eq. (2) we find MRa 2 = FR - fR = Ma = 2 F - 2 f (5) Comparing this formula with eq. (1), we see that 2 F - 2 f = F + f (6) and therefore f = F 3 = 1 . 43333 N . (7) Question 2, chap 13, sect 2. part 2 of 2 8 points Calculate the acceleration of the spool’s center of mass. Correct answer: 1 . 06173 m / s 2 (tolerance ± 1 %). Explanation: According to eqs. (1) and (7) Ma = F + f = 4 3 F (8) and therefore a = 4 F 3 M = 1 . 06173 m / s 2 . Question 3, chap 13, sect 2. part 1 of 1 8 points Consider a basketball rolling (without slip- ping) down an inclined plane. The basketball is basically a thin spherical shell, hence its moment of inertia I = 2 3 M R 2 . The acceleration of gravity is 9 . 8 m / s 2 . Given the angle θ = 32 . 6 between the incline and the horizontal, calculate the bas- ketball’s center’s acceleration. Hint: You don’t need the values of the basketball’s mass or radius. Correct answer: 3 . 16797 m / s 2 (tolerance ± 1 %). Explanation: Rolling motion of the basketball comprises the linear motion of the basketball’s center of mass (which is of course at the geometric center of the ball) and the rotation around the

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homework 11 – – Due: Apr 12 2007, 4:00 am 2 axis going through that center. The rotation governed by the equation I α = τ net , where the right hand side is the net torque of all the forces acting on the basketball. There are three such forces — the basket- ball’s weight vector W = M vectorg , the normal force vector N and the friction force vector f — but the vector W and the vector N act along lines going through the ball’s cen- ter and hence have zero lever arms. On the other hand, the lever arm of the friction force is the ball’s radius R , hence τ net = W × 0 + N × 0 + f × R , and consequently I α = f R . Now consider the linear motion of the bas- ketball’s center along the incline; by Newton’s Second Law, M a = F net along = M g sin θ - f .
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Kaplunovsky - hw 11 - homework 11 Due 4:00 am Question 1...

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