A uniform, solid cylinder with mass 2M and radius 2R rests on a horizontal table

Question

A uniform, solid cylinder with mass 2M and radius 2R rests on a horizontal tabletop. A massless string is attached to a massless yoke which is in turn attached to a frictionless axle that runs through the central axis of the cylinder, so that the cylinder can rotate about the axle. The string passes over a solid cylindrical pulley with mass M and radius R that is mounted on a frictionless axle. A block of mass M is suspended from the free end of the string. The string does not slip over the pulley surface, and the cylinder rolls without slipping on the tabletop. Use force and torque methods to derive an expression for the acceleration of the block after the system has been released from rest.

Explanation / Answer

If we imagine our block of mass M after having fallen a distance which I'll call Y, we can say the change in potential energy of the block is equal to all the translational and rotational kinetic energy of the system.

- The potential energy of the block is given by:
M·g·Y (g is acceleration to gravity)

- The total translational energy is equal to the sum of the translational kinetic energy of the center of mass of the cylinder of radius 2R plus the kinetic energy of the falling block:
1/2·M·V² + 1/2·M·V² = M·V²

where V is the velocity.

- The total rotational kinetic energy is equal to the sum of the rotational kinetic energy of the moving cylinder and

pulley. Rotational energy for an object = 1/2·I·², where I is the moment of inertia and is angular velocity.

For a cylinder the moment of inertiais 1/2·M·R², so remembering our pulley has radius R and our moving cylinder has radius 2R our total rotational energy is:

1/2·(1/2·M·R²)·² + 1/2·(1/2·M·(2R)²)·²

Note

is the angular velocity of the pulley and is the angular velocity of the cylinder.

- We can rewrite the angular velocity of the pulley, , in terms of V using v = r, hence:
= V/R

- We can rewrite the angular velocity of the cylinder, , in terms of V by knowing the relationship Vcm = r, where Vcm is the velocity of the center of mass of a cylinder. As our radius is 2R,

we find:
= V/2R

- Substituting the alternate angular velocity into our equation for rotational kinetic energy and simplifying we end up with:
1/2·M·V²

- Therefore the total energy when the translational and rotational energies are added is:
M·g·Y = 1/2·M·V² + M·V² = 3/2·M·V²

At this point we can note that our velocity, V, is really dY/dt, the rate of change of Y.

Hence our equation becomes:
M·g·Y = 3/2·M·(dY/dt)²

Cancelling M we get:
g·Y = 3/2·(dY/dt)²

Taking a derivative with respect to time and using the chain rule** we end up with:
g·(dY/dt) = 3·(d²Y/dt²)·(dY/dt)

Cancelling dY/dt we arrive at:
g = 3·(d²Y/dt²)

And

so

our acceleration of the block, (d²Y/dt²), is equal to g/3.


Note on the chain rule calculation:
d/dt·(dy/dt)² = 2·dy/dt·(d/dt·(dy/dt)) = 2·dy/dt·d²y/dt²

Note on why simply calculating the torque as M·g·R is wrong (other answers mistake):
The tension in the string from the pulley to the cylinder does not equal the tension in the string from the pulley to the block. Hence the force acting on the pulley is actually (Mg - T) where T is the tension from the pulley to the cylinder. So the torque on the pulley is (Mg - T)·R.

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