Help with a physics problem

Question

Answer 1

a.

- Each object is subject to downward force of gravity of magnitude mg.

- Each object is subject to tension T on its string in upward direction.

- The cylinder is subject to two tensions of magnitude T, but downward direction.

b., c.

Let a be the acceleration of the two objects.

(1) From Newton's second law applied to either object

mg - T = ma

(2) From Newton's second law applied to the cylinder

2TR = (1/2)MR²a/R

where

2TR is the total torque acting on the cylinder

(1/2)MR² is the moment of inertia of the cylinder

a/R is the angular acceleration of the cylinder

This equation (2) can be simplified into

4T = Ma

So we have two equations (1) and (2) and two unknowns -- T and a.

If we use the abbreviation

Q = m/(M+4m)

then the solution is

a = 4gQ

T = MgQ

The angular velocity starts with 0 and at time t will be at/R = 4gQt/R

d.

Angular acceleration is

a/R = 4gQ/R

e.

The objects drop distance h in time

t = √(2h/a)

In that time they reach velocity

v = at = √(2ha) = 2√(2hgQ/R)