Falling mass attached to wheel

[axe34]

Homework Statement

Homework Equations

I am aware that this can be done via energy methods. However, I wish to do it via:

integral between t1 and t2 of Torques about point (say centre of the wheel) = change in angular momentum of the wheel.

The Attempt at a Solution

The torque is mgR. t1 is zero.
I get that the block hits the floor at t = (square root of 2gh)/g

When I use the change in angular momentum equation, I get w (omega) final as (2.m. (square root of 2gh))/MR but this supposedly is not correct.

Any ideas??

 

[Orodruin]

I get that the block hits the floor at t = (square root of 2gh)/g

This cannot be correct. It is only correct when the cylinder is massless. Consider the case where the cylinder is much more massive than the attached mass - the acceleration will be much lower due to the torque needed to accelerate the rotation of the cylinder. Your result must depend on the ratio of masses. It seems to me that you are ignoring how the cylinder and hanging mass are affecting each other through the tension in the string.

 

[axe34]

I've looked into this a bit further - I was neglecting the possible tension in the cable. I thought that a non-stretching cable meant no tension but apparently not!

 

[Orodruin]

You do not need to compute the tension. If you consider the system including both cylinder and falling mass, the tension is internal and does not affect the angular momentum.

 

[axe34]

Yes, but the torque is not just mgR now, but probably (mg-T)R

 

[ehild]

I've looked into this a bit further - I was neglecting the possible tension in the cable. I thought that a non-stretching cable meant no tension but apparently not!

Non-stretching cable means that its length does not change.
There is tension in the cable. That tension acts both on the cylinder and the falling mass.

 

[Orodruin]

Yes, but the torque is not just mgR now, but probably (mg-T)R

This is because you are now considering the cylinder only as your system. If you consider the cylinder and hanging mass as one system as I suggested, then:

If you consider the system including both cylinder and falling mass, the tension is internal and does not affect the angular momentum.

The torque on the full system (including both cylinder and hanging mass) is definitely still mgR. What you have to figure out is what the angular momentum is.