This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Morgan, Mena Homework 11 Due: Nov 19 2007, 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 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 = 42 between the incline and the horizontal, calculate the basketballs centers acceleration. Correct answer: 3 . 93449 m / s 2 . Explanation: Rolling motion of the basketball comprises the linear motion of the basketballs center of mass (which is of course at the geometric center of the ball) and the rotation around the axis going through that center. The rotation governed by the equation I = 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- balls weight ~ W = M ~g , the normal force ~ N and the friction force ~ f but the ~ W and the ~ N act along lines going through the balls cen- ter and hence have zero lever arms. On the other hand, the lever arm of the friction force is the balls radius R , hence net = W (0) + N (0) + f R , and consequently I = f R . Now consider the linear motion of the bas- ketballs center along the incline; by Newtons Second Law, M a = F net along = M g sin - f . Since the ball rolls down without slipping, its linear motion is related to its rotation according to v = R and hence a = R . Consequently, I R 2 a = I R = f , and therefore M + I R 2 a = ( M g sin - f ) + f = M g sin . For a thin spherical shell such as a basket- ball, M + I R 2 = M + 2 3 M = 5 3 M . Consequently, 5 3 M a = M g sin and there- fore a = 3 5 g sin = 3 5 (9 . 8 m / s 2 )sin42 = 3 . 93449 m / s 2 . keywords: 002 (part 1 of 1) 10 points Consider the problem of the solid sphere rolling down an incline without slipping. The incline has an angle . The spheres length up the incline is , and its height is h . At the beginning, the sphere rests on the very top of the incline. The sphere has mass M and radius R . The acceleration of gravity is 9 . 8 m / s 2 . Hint: The moment of inertia of a sphere with respect to an axis through its center is 2 5 M R 2 ....
View Full Document