A pulley has a light string passing over it one end of which

A pulley has a light string passing over it, one end of which carries a 5kg mass, and the other end, hanging over the other side of the pulley carries a mass of 2kg. If the pulley has a moment of inertia of 1.25kgm^2 adn a groove diameter of 0.8m, and the bearings exert a friction torque of 1.5N*m, calculate the angular acceleration of the pulley, assuming the string does not slip.

Solution

The net torque trying to turn the pulley would be due to the masses hanging on the either side of it.

Let us assume the angular acceleration of the pulley be A, hence the linear acceleration of the masses would be Ar

where r is the radius of the pulley, in this case, equal to 0.4 m

There for the mass we have: 5g - T1 = 5*0.4 * A

For the other mass we have T2 - 2g = 2*0.4*A

Now, for the pulley we have: (T1 - T2) * 0.4 - 1.5= 1.25*A

We add the first two equations to get: 3g + T2 - T1 = 2.8 A

That is T1 - T2 = 3g - 2.8A

using this in equation 3 we get:

(3g - 2.8A) * 0.4 - 1.5 = 1.25 A

1.2g - 1.5 = 2.37 A

Hence A = 4.334

NOTE: The tensions to the left and right of the pulley are different as the pulley is not massless and hence the tension will not get transmitted without change.