Question

A string is wrapped around a uniform disk of mass M and radius R. Attached to

Answer 1

Answer:

Change in translational KE is given as

$\Delta k = FL + (4m + M)gh - \frac{1}{4}MR^2\omega^2 - 2mb^2\omega^2$

Explanation:

As we know by work energy theorem that work done by all forces is equal to the change in kinetic energy of the system

So here we know that there are two forces acting on the system

So work done by the system of this force is equal to the change in kinetic energy

So we have

$W = FL + (4m + M)gh$

So we have

$FL + (4m + M)gh = \Delta K + \frac{1}{2}(\frac{1}{2}MR^2 + 4mb^2)\omega^2$

so we have

$\Delta k = FL + (4m + M)gh - \frac{1}{4}MR^2\omega^2 - 2mb^2\omega^2$