Solid pulley & mass on an inclined plane

[Ampere]

Homework Statement

See the image.

Homework Equations

T=Iα, F=ma, I=0.5*m*r^2, Friction=u*N

The Attempt at a Solution

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

m_blockgsin(30)-Tension-Friction=m_block*a

so

m_blockgsin(30)-m_cylinder*a/2-μm_blockgcos(30)=m_block*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.

 

[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.

 

[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.