# Solid pulley & mass on an inclined plane

1. Jun 25, 2013

### Ampere

1. The problem statement, all variables and given/known data

See the image.

2. Relevant equations

T=Iα, F=ma, I=0.5*m*r^2, Friction=u*N

3. The attempt at a solution

I used the torque equation to get Tension = mcylinder*a/2, and then plugged that into a force equation for the block.

mblockgsin(30)-Tension-Friction=mblock*a

so

mblockgsin(30)-mcylinder*a/2-μmblockgcos(30)=mblock*a

I ended up with μ=0.36. Just wanted to check my answer because I never ended up using the radius of the cylinder (it canceled out), so I was wondering why it was given.

#### Attached Files:

• ###### problem.jpg
File size:
27.6 KB
Views:
266
Last edited: Jun 25, 2013
2. Jun 26, 2013

### thecommexokid

I agree with you.

I agree with you. The torque on the pulley is supplied by the tension, so
$\tau = T \times R_{pulley}$.​
A point along the circumference must have a linear acceleration equal to the acceleration of the block, so
$\alpha_{pulley} R_{pulley} = a_{block}$.​
And the moment of inertia for a solid cylinder is
$I=\frac12 MR^2$​
So
$\tau = I\alpha = \left(\frac12 M R^2\right)\left(\frac a R\right) = T \times R$.​
If we solve for the tension, all the R's indeed drop out:
$T=\frac12 M a$,​
exactly like you found.

Last edited: Jun 26, 2013
3. Jun 26, 2013

### Ampere

Yup. I just did this problem again using energy conservation - if the acceleration is 1.6 m/s^2, then in 1 second the block gains 1.6m/s of speed and travels 0.8m while the pulley attains an angular speed of 32 rad/s. Energy conservation gives me the same answer for u so maybe that's why the radius was given, in case you wanted to solve it that way.

Thanks.