# Rotational Dynamics: Tension

1. Jan 2, 2014

### henryli78

1. The problem statement, all variables and given/known data
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. 6.9 N
B. 9.8 N
C. 16 N
D. 26 N
E. 29 N

2. Relevant equations
∑$\tau$ = I$\alpha$
∑F = ma
a = $\alpha$r

3. The attempt at a solution
Honestly I'm not very sure how to begin. The two upwards acting forces seem to be the Tension forces of the string + suspension, while the downwards forces seem to be the weights of the objects. I am unsure how to relate the rotational inertia of the disk/cylinder with the tension forces, seeing that we don't know the net torque of the system (no angular acceleration).

Can someone direct me and maybe give me a hint on where to start? Much appreciated.

Last edited: Jan 2, 2014
2. Jan 2, 2014

### tiny-tim

hi henryli78!

draw two free body diagrams, one for the mass and one for the disk (cylinder)

do F = ma for the mass, and F = ma and τ = Iα for the disk …

show us what you get

3. Jan 2, 2014

### CWatters

Hint: The 2kg object is accelerating downwards pulling on the string and spinning up the cylinder.

Not if the 2kg object is accelerating downwards.

4. Jan 2, 2014

### henryli78

Ok I get:
For the mass: m_1*g - T_1 = m_1*a
For the disk: mg + T_1 = T_2
I*alpha = T_1*r

Is this ok so far?

5. Jan 2, 2014

### haruspex

Yes. you need one more equation. Have you got it?

6. Jan 2, 2014

### henryli78

Do you mean alpha = a/r?

7. Jan 3, 2014

Yes...

8. Jan 3, 2014

### Malverin

Here you can see how to solve this in principal.
Yo-yo has not the same rotational inertia, but you can see there what equations to use to solve this.