You have volunteered to help organize a 70\'s party for engineering students at

Question

You have volunteered to help organize a 70's party for engineering students at LSU. At the last minute, you realize the spinning mechanism for the disco ball is broken, and you decide to improvise. You wrap a thin string many times around the support pivot of the disco ball, run the string over a pulley, and then attach the string to an aluminum travel coffee mug. The setup is shown in the figure, and the technical specifications are listed below. Technical specifications: The support pivot has radius rs and is effectively massless. The disco ball is more or less spherical, and its mass is primarily due to the thin glass tiles on its surface. The glass tiles have a total mass M, and the disco ball has an average radius R. The pulley has radius rp and a moment of inertia Ip about its frictionless axle. The coffee mug has mass m. The string is massless and does not stretch. The string is wrapped so that the disco ball spins counterclockwise about the +y axis as the string unwinds. Draw free body diagrams for the disco ball, the pulley, and the coffee mug. (The smaller circle for the disco ball represents the support pivot, and the arrows indicate the rotation direction as the string unwinds.) Be sure to label all forces and any relevant radii. What is am, the magnitude of the linear acceleration of the mug? Report your answer using m, M, R, rp, lp, and numerical and physical constants as needed. Assume that the mug is released from a height h above the ground while all three objects are at rest. Using energy methods (and not kinematics), determine the speed ty with which the coffee mug hits the ground. Report your answer using m, M, R, rs, rp, lp, h, and numerical and physical constants as needed.

Explanation / Answer

Let tension in string = T

mg*r_p - T*r_p = I_p*alpha...........where alpha = angular acceleration of pulley

alpha = a_m / r_p

Therefore, mg*r_p - T*r_p = I_p* a_m / r_p................1

For sphere, inertia I = 2/3*MR^2

For ball, Torque = I*alpha = I*a_m/r_s

T*r_s = 2/3*MR^2 *a_m/r_s

Thus, T = 2/3*MR^2 *a_m/r_s^2

Putting it in eqn 1:

mg*r_p - (2/3*MR^2 *a_m/r_s^2)*r_p = I_p* a_m / r_p

mg*r_p = a_m*[2/3*MR^2 *r_p/r_s^2 - I_p / r_p]

a_m = mg*r_p / [2/3*MR^2 *r_p/r_s^2 - I_p / r_p]

c)

Potential energy = m*a_m*h

Kinetic energy = 1/2*m*v_f^2

m*a_m*h = 1/2*m*v_f^2

m*mg*r_p / [2/3*MR^2 *r_p/r_s^2 - I_p / r_p]*h = 1/2*m*v_f^2

v_f = sqrt [2*mg*r_p / [2/3*MR^2 *r_p/r_s^2 - I_p / r_p]*h]

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