Yo-Yo man releases a yo-yo from rest and allows it to drop, as he keeps the top

Question

Yo-Yo man releases a yo-yo from rest and allows it to drop, as he keeps the top end of the string stationary, as shown in the image. The mass of the yo-yo is 0.056kg, its moment of inertia is 2.9 *10-5kg.m2, and the radius, r, of the axle the string wraps around is 0.0064m.

a) What is the linear speed, v, of the yo-yo after it has dropped through a height 0.50hm

b) If the yo-yo%u2019 moment of inertia is increased, does its final speed increase, decrease, or stay the same? Justify your answer. Calculate the final speed for the case I = 3.9*10-5kg.m

was asked before but not answered,i have the first part answered but not part b)

Explanation / Answer

First let us find the angular velocity of the yo-yo. The string applies a torque to the axle which causes the yo-yo to rotate.

KE = linear kinetic energy of the yo-yo = (1/2)M*V^2

KEr = rotational kinetic energy = (1/2)*I*w^2 = (1/2)*I*(V/R)^2

PE = gravitational potential energy = M*g*H

PE = KE + KEr

M*g*H = (1/2)M*V^2 + (1/2)*I*(V/R)^2

M*g*H = (1/2)*V^2{M + I/R^2}

H = (1/2)*V^2{M + I/R^2} / (M*g)

H = (1/2)(0.55)^2 {0.056 + 2.9x10^(-5)/(0.0064)^2}/(0.056*9.8)

H = 0.21 m

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