Question

A uniform cylinder of mass 100 kg and radius 50 cm is mounted so it is free to rotate about a fixed, horizontal axis that passes through the centers of its circular ends. A 10-kg block is hung from a massless cord that is wrapped around the cylinder’s circumference. When the block is released, the cord unwinds and the block accelerates downward. a) What is the block’s acceleration?

Answer 1

The equation of motion for the falling block (mass m) is (tension in the cord is T):

m a = m g - T

The equation for the rotating cylinder (mass M, radius R, angular acceleration A):

1/2 M R^2 A = T R

because 1/2MR^2 is the moment of inertia for a solid cylinder about its axes and T R is the external torque.

When no slip occurs then R A = a, so this last equation becomes

1/2 M a = T


So now we have two equations for the two unknowns a and T:

ma = m g - T

1/2 Ma = T

Substituting the second expression for T in the first allows us to obtain the acceleration of the block, a:

m a = m g - 1/2 M a

(m + M/2) a = m g

a = m g /(m + M/2)

a = g/(1 + M/(2m))

a = 9.81 m/s^2 / ( 1 + 100kg/(2*10kg))

a = 1.64 ms^2