# Homework Help: A mass attached to a pulley- torque

1. Jul 18, 2013

### natasha13100

Could someone please explain to me where I am going wrong? I've tried combining the relevant equations below in multiple ways. Also, I'm not sure if my answer is given in rad/s2 or °/s2.
1. The problem statement, all variables and given/known data
A uniform solid disc of mass 1.2 kg and radius 0.24 m is free to rotate on a horizontal frictionless axle passing through the center of the disc. A long light string is wrapped around the disc; a block of mass 1.9 kg is suspended on the string as shown. The system is released from rest, and the block begins to descend. Find the magnitude of the angular acceleration α of the disc. Enter your answer in rad/s2 and use g = 9.8 m/s2.

2. Relevant equations
force(F)=mass(m)*acceleration(a)
torque(t)=radius(R)*F*sin(θ) where θ=angle between R and F when the tails are placed together
Ʃt=moment of inertia(I)*angular acceleration(α)
I=mR2
tangential acceleration(atan)=Rα
I=1/2mR2/SUP] for discs of uniform density

3. The attempt at a solution
I use m for the mass of the block and M for the mass of the disc. a is the tangential acceleration/acceleration of the block and α is the angular acceleration.
For the disc, there is a normal force exerted by the axle that is equal in magnitude and opposite in direction to the force of gravity. There is also string tension(T).
Two forces are acting on the mass: string tension and gravitational force.
The two equations relevant to α are Ʃt=I*α and a=R*α.
All of the torque is coming from the force T so Ʃt=t=RFsin(θ)=.24T. Additionally, I=1/2MR2=1/2(1.2)(.24)2=.03456. Therefore, .24T=.03456α and T=.144α
Tension also acts on the mass.
The net force(F)=T-the force due to gravity(G)=T-mg and F=ma so ma=T-mg and T=ma+mg=1.9a+1.9*9.8=1.9a+18.62
Therefore, .144α=1.9a+18.62. Next, I use a=Rα to make the equation .144α=1.9(.24)α+18.62 or .144α=.456α+18.62.
.312α=-18.62 so α=-4655/78≈-59.67948718

2. Jul 18, 2013

### TSny

The normal force is not equal to the force of gravity. (There are three vertical forces acting on the disk that must add to zero.) However, this doesn't affect your solution since the normal force doesn't create a torque.

In your calculations, you need to be careful with positive and negative directions. Does your positive direction for the angular acceleration of the disk correspond to your positive direction for the acceleration of the block?

3. Jul 18, 2013

### natasha13100

I used the right hand rule to check and it doesn't. Therefore, α is positive 59.67948718.

4. Jul 18, 2013

### TSny

You need to consider if the relation between the acceleration of the block and the angular acceleration of the pulley is a = +αR or a = -αR.

When the block is released, the pulley will start to undergo angular acceleration in a particular direction (CW or CCW depending on which side the block is on). Suppose you take positive angular acceleration to correspond to the direction the pulley accelerates.

The block will accelerate downward. So, it would be convenient to take downward as the positive direction of acceleration for the block. That way, when α is positive then a will also be positive and you can write a = +αR.

But if you take downward as the positive direction for the block, how should you choose the signs of the forces when you set up ƩF = ma for the block?

5. Jul 19, 2013

### natasha13100

ƩF=mg-T=ma

6. Jul 19, 2013

### TSny

Right. So, go back to your calculations and use this equation instead of T - mg = ma

7. Jul 20, 2013

### natasha13100

The net force(F)=the force due to gravity(G)-T=mg-T and F=ma so ma=mg-T and T=mg-ma=1.9*9.8-1.9a=18.62-1.9a
Therefore, .144α=18.62-1.9a. Next, I use a=Rα to make the equation .144α=-1.9(.24)α+18.62 or .144α=-.456α+18.62.
.6α=18.62 so α≈31.03333333333333

8. Jul 20, 2013

### TSny

That looks correct to me. The equation a = αR requires the rotation angle to be in radians. So, your answer for α will be rad/s2.