Calculating Speed of Descending Block B using Energy Methods

[~angel~]

Ok...I don't have a picture for this but imagine a block A resting on a table. It is attached to a string which in turn is attached to a pulley on the edge of the table. This same string is attached to another block B, which is vertically hanging off the pulley.

The pulley has radius R and moment of inertia I. The rope does not slip over the pulley, and the pulley spins on a frictionless axle. The coefficient of kinetic friction between the table top and block A is mu_k. The system is released from rest, and block B descends. Block A has mass m_A and block B has mass m_B. Use energy methods to calculate the speed of block B as a function of the distance that it had descended.

I'm not sure how to incorporate hte moment of inertia and the rotational motion of the pulley.

Could someone please help.

Thank you.

 

[SpaceTiger]

I'm not sure how to incorporate hte moment of inertia and the rotational motion of the pulley.

The rotational energy of an object with moment of inertia, I, is just:

$$E=\frac{1}{2}I\omega^2$$

What are the other energies in the problem? What are their initial and final states?

 

[~angel~]

Initial KE is zero. I'm not sure about PE. Say that we define the zero point to be the distance d block B has descended. Then it'll have PE m_B+g+d initially. Ihave no idea about the PE for block A, but coiuld it remain the same?

Final KE include (m_A*v_A^2)/2, (m_B*v_B^2)/2 and I*w^2/2. There is also friction...The work done by that is mu_k*m*g*d. Final PE is zero.

I'm not sure though.

 

[SpaceTiger]

That all looks right, assuming you meant the initial PE was m_B*g*d. Now just set them equal and solve.

 

[~angel~]

Yeah, i meant that. So what exactly is the PE of block A...would that be m_A*g*d and remain the same during each state?

 

[nishant]

p.e of block A will remain same except for the fact if you want it to get off the surface on which it is resting

 

[~angel~]

Ive got the answer. Thank you.