Newton's Second Law and Rotation

[whoareyou]

Homework Statement

A 0.70-kg disk with a rotational inertia given by MR^2/2 is free to rotate on a fixed horizontal
axis suspended from the ceiling. A string is wrapped around the disk and a 2.0-kg mass hangs
from the free end. If the string does not slip, then as the mass falls and the cylinder rotates,
the suspension holding the cylinder pulls up on the cylinder with a force of:
A. 6.9N
B. 9.8N
C. 16N
D. 26N
E. 29N

Homework Equations

Newton's second law for rotating bodies

The Attempt at a Solution

I was able to get the acceleration of the disk + the mass by myself following the solution below. It's probably pretty obvious but I just can't seem to figure out how he got T = mg + M(g-at), specifically the M(g-at) part assuming that the suspension holding the cylinder pulls up on the cylinder with a force equal to the sum of the weights of the pulley and the block.

This is the solution I am trying to make sense of:

The suspension holding the cylinder pulls up on the cylinder with a force of

T = mg + M(g-at), where m - disk, M - 2kg mass, at - the acceleration of a 2.0-kg mass. At the same time that is the tangential acceleration of the disk. Let's find it.

J dω/dt = M(g-at) * R,

mR²/2 * 1/R dV/dt = M(g-at) * R, where at dV/dt = at.

m at/2 = M(g-at)

at = 2g/(m/M+2) = 2*9.8/(0.7/2 + 2) = 8.34 m/s²

T = mg + M(g-at) = 0.7*9.8 + 2*(9.8-8.34) = 6.86 + 2*1.46 = 9.8 N

 

[Andrew Mason]

Homework Statement

A 0.70-kg disk with a rotational inertia given by MR^2/2 is free to rotate on a fixed horizontal
axis suspended from the ceiling. A string is wrapped around the disk and a 2.0-kg mass hangs
from the free end. If the string does not slip, then as the mass falls and the cylinder rotates,
the suspension holding the cylinder pulls up on the cylinder with a force of:
A. 6.9N
B. 9.8N
C. 16N
D. 26N
E. 29N

Homework Equations

Newton's second law for rotating bodies

The Attempt at a Solution

I was able to get the acceleration of the disk + the mass by myself following the solution below. It's probably pretty obvious but I just can't seem to figure out how he got T = mg + M(g-at), specifically the M(g-at) part assuming that the suspension holding the cylinder pulls up on the cylinder with a force equal to the sum of the weights of the pulley and the block.

Draw free body diagrams. The forces acting on the axle must sum to 0 since the axle is not accelerating. The forces acting on the falling mass M must add to Ma_t.

The forces acting on the axle are the tension in the axle support, T_x (upward), mg (downward) and the tension in the string (downward). The forces acting on the falling mass are: Mg (downward) and the tension in the string, T_s(upward).

So:

T_x - mg - T_s = 0

and

Mg - T_s = Ma_t

These reduce to:

T_x = mg + Mg - Ma_t

AM

 

[whoareyou]

The forces acting on the axle must sum to 0 since the axle is not accelerating.

By axle, do you mean the disk, or the thing that's supporting the disk+mass to the ceiling? Also (assuming by axle you mean the disk) how come it isn't accelerating? Isn't there a net torque caused by the tension in the string causing angular acceleration in the disk?

 

[Andrew Mason]

By axle, do you mean the disk, or the thing that's supporting the disk+mass to the ceiling? Also (assuming by axle you mean the disk) how come it isn't accelerating? Isn't there a net torque caused by the tension in the string causing angular acceleration in the disk?

You state that the disk is free to rotate about a fixed axis suspended from the ceiling. I appeared to me that this axis passed through the centre of the disk (radius R). So it appears to me that the disk rotates on a fixed horizontal axle through its centre of mass. How else would it rotate?

The centre of mass of the disk does not accelerate because the axle is fixed. What you want to determine is the upward force on that axle.

AM

 

[CWatters]

how come it isn't accelerating? Isn't there a net torque caused by the tension in the string causing angular acceleration in the disk?

Andrew means it doesn't have linear acceleration (it does have angular acceleration)

I just can't seem to figure out how he got T = mg + M(g-at), specifically the M(g-at) part assuming that the suspension holding the cylinder pulls up on the cylinder with a force equal to the sum of the weights of the pulley and the block.

Your assumption is incorrect or perhaps you need to think about the "weight of the block M". The mass M is accelerating downwards so the suspension isn't carrying it's full weight (eg it's not Mg).

Consider..

If the disc was locked so that mass M was just hanging there then the force on mass M would be Mg and the force on the suspension would be mg + Mg.

If mass M was somehow in total free fall (eg the string had been cut) then then force due to mass M would be zero. In that case the force on the suspension would be just mg + 0

So in your problem the force on the suspension is going to be somewhere between these two.

What is the tension in the string?

 

[whoareyou]

OK, so to summarize:
- the disk has angular acceleration and the block has linear acceleration
- the disk has no linear acceleration, so by Newton's Second Law (not for rotation), it has no (linear) acceleration so the sum of the forces = 0
- the tension in the string causes a torque on the disk so it has the same tangential acceleration as the hanging mass

Hmm, I think those are the key ideas to take away. Am I missing something. I think in the end, once I realized that why the disk is not accelerating, the whole solution became much more clear.