# Finding moments of pulleys with mass and tensions.

1. Mar 22, 2008

### jrrodri7

A uniform Disk of radius 0.18 m is mounted on a frictionless, horizontal axis. A light cord wrapped around the disk supports a 0.57 kg object. When released from rest, the object falls with a downward acceleration of 2.6 m/s^2

What is the Moment of Inertia of the disk?

[Broken]

Relevant Equations
$$\Sigma$$ F = ma

$$\Sigma$$ $$\tau$$ = I $$\alpha$$

$$\alpha$$ = a / r

My attempt

The Cord is massless and the Axis is frictionless, so that all equals to 0.
I set F_x = 0, and F_y = T + H - mg, T is the Tension force the pulley is pulling away from the weight attached to it by the string. So i defined the T as "Mg" where M is the mass of the pulley and "mg", where m is 0.57 kg. Finding the Tension yielding 5.586 N.

I set the sum of the torques equal to I$$\alpha$$, and substituted alpha as a/r. given the a causing rotational motion, because the system is being essentially rotated from the weight falling from rest attached to the pulley, making the 2.6 = a, and the r = 0.18 m. Given the substitution for $$\alpha$$, I can substitute the sums of torques as the magnitude of torques yielding rFsin(theta), since it's 90 degrees, the sin becomes 1, making the equation essentially F * radius. So, given the T from prior, 5.586, i multiplied by 0.18 m, attating 1.00548. I set 1.00548 yielding I x (2.6/0.18), and I solved for I, giving me a number of 0.069627. This is obviously incorrect though. I've been having trouble understanding the values of the application of how to SEE these torques in pictures and such. It's been giving me ALOT of trouble, so any help would be useful.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: Apr 23, 2017 at 11:42 AM
2. Mar 23, 2008

### Staff: Mentor

Find the tension by analyzing the forces acting on the hanging mass. You have the acceleration.

Once you correctly find the tension, use it to find the torque acting on the disk.

3. Mar 23, 2008

### jrrodri7

Oh snap! Since I have 2 separate accelerations, all I'd have to do is find the difference in those forces between g and the 2.6 m/s^2! It'd end up being something like,

T = mg - ma_y

then set T = Mx, x being the difference's in acceleration.

giving Mx = mg - ma (all in y directions).

M(9.8-2.6) = (.57 * 9.8) * (.57 * 2.6)

7.2M = 5.586 * 1.482

7.2M = 8.28

M= 1.15 kg, which makes sense since it should be heavier then the weight.

Then I apply (1/2) M R ^2. and I arrive at, I = 0.01863

4. Mar 23, 2008

### Staff: Mentor

This is all you need to solve for T, since m and a_y are given.
No idea what you're doing here.

5. Mar 24, 2008

### jrrodri7

oh, I was just saying that the only force pulling against the weight would be the pulley itself, and that it's mass and w/e acceleration it would deduct from the prior acceleration. But yes, it's only the first set..okay.

6. Mar 24, 2008

### jrrodri7

Then I'd have find the Force on the pulley as well correct?

F_y (pulley) = T - mg, and with that since it's not moving and no friction, i can set it to 0, and find that the only tension on the pulley to the horizontal is Mg, I don't know what M is though...would i have to substitute T i find from the other equation and find it?

7. Mar 25, 2008

### Staff: Mentor

Once you've found the tension, you'll need to find the torque it exerts on the pulley. Note that the string is the only thing that exerts a torque on the pulley.