In the diagram below, the pulley has a radius R and a moment of inertia I. The r

Question

In the diagram below, the pulley has a radius R and a moment of inertia I. The rope does not slip over the pulley, and the pulley spins on a friction less axle. The coefficient of kinetic friction between the block and the table top is muk. The block on the table top has mass m_1. The descending block has mass m_2. The system is released from rest. Find: The speed v and the acceleration a of the blocks when the distance of descent is d The tension forces which the rope exerts on the blocks. Which of the two tensions is larger?

Explanation / Answer

Let tension in the rope attached to m2 be T2
Let tension in the rope attached to m1 be T1

Let acceleration of the system be a
Force balance on m2 => m2g - T2 = m2a
Force balance on m1 => T1 - mu*m1g = m1a
Torque balance on pulley => (T1 - T2)R = I (alpha)
but alpha = a/R
so, (T1 - T2)R = Ia/R

so, m2g - mu*m1g + Ia/R^2 = a(m2 + m1)
a) g(m2 - mu*m1)/(I/R^2 + m2 + m1) = a
after descent d, 2*a*d = v^2
v = sqroot(2d*g(m2 - mu*m1)/(m1 + m2 + I/R^2)
b) T1 = m1*g(m2 - mu*m1)/(m1 + m2 + I/R^2) + mu*m1g = m1*g[m2 - mu*m1 + m1*mu + m2*mu + I*mu/R^2]/[m1 + m2 + I/R^2] = m1*g[m2 + m2*mu + I*mu/R^2]/[m1 + m2 + I/R^2]
T2 = m2(g - a) = m2g(1 - (m2 - mu*m1)/(m1 + m2 + I/r^2)) = m2*g(m1 + m2 + I/R^2 - m2 + mu*m1)/(m1 + m2 + I/R^2) = m2*g(m1 + mu*m1 + I/R^2)/(m1 + m2 + I/R^2)
c) T1 - T2 = g[m2m1 + m1m2mu + Im1mu/R^2 - m1m2 - mum1m2 - Im2/R^2]/(m1 + m2 + I/R^2) = dI[m1*mu - m2]/R^2(m1 + m2 + I/R^2) > 0 [ as the blocks move downwards)

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