Question

The figure shows two masses connected by a cord passing over a pulley of radius R₀ and moment of inertia I. Mass M₁slides on a frictionless surface, and M₂ hangs freely.

(a) Determine a formula for the angular momentum of the system about the pulley axis, as a function of the speed v of mass M₁ or M₂. (Use R0 for R₀, M1 for M₁, M2 for M₂, I,v and g for gravity, as necessary.)

(b) Determine a formula for the acceleration of the masses.

Answer 1

The angular momentum, about some point, of a mass moving in a straight line a perpendcular distance "r" from its line of motion to the point is mvr.
So in your problem the line of motion of each mass is a perpendicular distance "R" from the center of the pulley and each mass then has angular momentum MavR & MbvR (equal velocities because they're tied together). And the pulley itself has angular momentum Iw or Iv/R.
So the total is;
L = MavR + MbvR + Iv/R = [(Ma+Mb)R + I/R]v

In second part, look at the forces on each mass separately and apply 2nd Law;
On Ma only horizontal force is string tension (take + in direction of acceleration)
Ta = MaA (I used "A" for acceleration)

On Mb two vertical forces upward string tension & downward weight;
Mbg - Tb = MbA

Next, apply the torque eq to the pulley. The torques are due to the string tension, TaR a cclw torque and TbR a clw torque;

TbR - TaR = I(alpha) = IA/R

In this last eq, elliminate Ta & Tb from the force eqs on each mass, then solve for "A".
That should get it.

[(M(a) +M(b))*R(0) +(1/R(0))]*v for part a.)

(M(b)*g)/(M(a) +M(b)+(1/R(0)^2)) for part b.)