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Two blocks of mass m and 3m are slug over a pulley by a massless rope. The massess are initially at the same height. The pulley is a uniform solid disk of mass 12m and radius R. The total length of the rope including the part that is slung over the pullley is L The masses are released at time t = 0
Determine the time it takes for the pulley to complete one revolution
Determine the time that it takes for the smaller mass to crash into the pulley.

https://brainmass.com/physics/torques/time-takes-pulley-complete-one-revolution-11436

SOLUTION This solution is FREE courtesy of BrainMass!

Torque acting on the pulley= (3m-m) x g x R= 2 m g R ----(1)
Torque= I alpha

Where I = moment of inertia = ½ M R ^2
And alpha = angular acceleration
M= mass of the pulley = 12 m
Therefore I= ½ x (12 m ) R ^2= 6 m R^2

Torque= I alpha = 6 m R^2 x alpha ----------(2)
Equating 1 and 2
6 m R^2 alpha = 2 m g R

or alpha = g/(3 R)

Determine the time it takes for the pulley to complete one revolution

theta = w o t + ½ alpha t ^2
But w o= 0
theta = 2 pi for 1 revolution
2 pi =½ alpha t ^2=1/2 (g/(3 R)) t ^2
or t^2= (12 pi R/g)
or t=square root of (12 pi R/g)

Determine the time that it takes for the smaller mass to crash into the pulley

Distance to be traveled for the smaller mass to crash into the pulley= (L-pi R)/2
(length of rope wound around the pulley is pi R)

This corresponds to ((L-pi R)/2) / 2 pi R = (L-pi R)/ 4 pi R revolutions

Angular displacement corresponding to (L-pi R)/ 4 pi R revolutions
=( (L-pi R)/ 4 pi R) *(2pi )= (L-pi R)/2R

theta = w o t + ½ alpha t ^2
But w o= 0
theta = (L-pi R)/2R
(L-pi R)/2R =½ alpha t ^2=1/2 (g/(3 R)) t ^2
or t^2= 3(L- pi R)/g
or t=square root of 3(L- pi R)/g

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!