Question

An green hoop with mass m_h = 2.8 kg and radius R_h = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass m_d = 2.4 kg and radius R_d = 0.08 m. The other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass m_s = 3.4 kg and radius R_s= 0.19 m. The system is released from rest.

1) What is the magnitude of the angular acceleration of the disk pulley?

2) What is the magnitude of the angular acceleration of the sphere?

Answer 1

In thie given situation, the only force on the system is the weight of the hoop

F_net = m g = ( 2.8kg ) ( 9.81m/s^2) = 27.468 N

In this case, the inert masses that must be accelerated are the ring, pulley, and the rolling sphere.

First, find the mass equivalent of M the pulley:

torque, τ = F*R = I*α = I*a/R

From Newton's second law. F = M*a = I*a/R^2

Rearrange the equation for mass equvilent of pulley

M = I/R^2 = ( 21/2*m*R^2 ) /R^2 = 1/2*m

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Mass equivalent of the rolling sphere:

From the parallel axes theorem, the moment of inertia of the sphere

I = 2/5 ( mR^2 ) ( for rotation about the center of mass) + mR^2 ( for the distance of the axis of rotation from the center of mass of the sphere )

I = 7/5*mR^2

Thus, the mass equivalent of the rolling sphere

M = 7/5*m

Find the acceleration

a = F/m = 27.468 N /(2.8 kg + 1/2*2.4 + 7/5* 3.4)

a = 3.1356 m/s^2

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( 1 )

Angular acceleration disk pulley:

α = a /Rd = 3.1356 m/s^2 / 0.08 m = 39.19 rad/s^2

(2 )

Angular acceleration sphere

α = a/Rs = 3.1356 m/s^2 / 0.19 m = 16.5 rad/s^2