# Pulley with Cylinder

1. Dec 16, 2003

### SuperGeek

I am stuck on this rotational motion with pulley question. Any help would be appreciated.

A block M2=5kg hangs off a pulley with mass from a string with tension T2 over a table edge.
A cylinder M1=3kg on the horizontal table top is attached to the string on the other end of the pulley (there is friction between the cylinder and table).
The cylinder is allowed to roll while the block is allowed to fall.
I need to compute linear acceleration of the roller-pulley-block system. The problem I have is that I am not sure what to use as the horizontal force on the cylinder.

Forces on m2: M2g - T2 = M2a
Forces on pulley: T2-T1 = 1\2ma
Forces on m1: N - m1g = 0 (vertical axis)
T1 - f = m1a
Should f = (Inertia of center of mass \ R^2)(a) ?

Thanks in advance for any responses.

2. Dec 17, 2003

### harsh

I am little confused about your problem. The pulley with mass is also a cylinder? Since it has mass, it must rotate.

3. Dec 17, 2003

### jamesrc

Unless you've left something out of the problem statement, it doesn't seem like you have to consider inertial effects of the pulley (I certainly doubt the pulley is moving, as your equations seem to suggest). If you want to use a Newtonian approach (as opposed to an energy-based method) as you have started, you're first equation is fine:
$m_2a = m_2g - T$

where T is the tension in the string and you have taken a>0 to be a downward acceleration of m2.

Assuming the string is inextensible, there is only one acceleration to deal with in this problem. Also, with the assumption of a massless, frictionless pulley, you can say that the tension on either end of the pulley is the same. This gives your x-direction equation for the cylinder to be:
$m_1a = T - f$

You still need another equation, so look at a torque balance for the cylinder. (Another assumption here: the string loops around a groove or bearing in the cylinder (like a yo-yo, maybe) so that the cylinder can roll without slip while being pulled; there's no unwinding string to worry about.) Look at the sum of torques about the center of the cylinder:
$I\alpha = fR$
But we also know how angular acceleration relates to linear acceleration:
$\alpha = \frac{a}{R}$
(watch your sign conventions; I believe what I've typed above is consistent...)

Now you can put it all together, along with I = .5*m1R2, and solve for a.