# Finding angular acceleration of pulley attached to a mass

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1. Dec 3, 2015

### vetgirl1990

1. The problem statement, all variables and given/known data
A block of mass 2kg (mb) can slide down a frictionless 53 degree inclune, but it is connected to a pulley of mass 4kg (mp with a radius of 0.5m. The pulley may be treated as a disk.
What is the angular acceleration of the pulley?

2. Relevant equations
a = rα

3. The attempt at a solution
I first expressed each object in terms of Newton's 2nd Law
BLOCK:
ΣFx = mbsinθ - T = mba
ΣFy = N - mgsinθ = 0

PULLEY:
ΣFx = T = mmpa
ΣFy = N-mg = 0

Then, isolated for T (tension) for both the block and pulley and equated them to each other. Their tension would be the same because the pulley and block are connected by the same rope.
(1) T = mbgsinθ - mba
(2) T =mpa
mpa = mbgsinθ - mba
a = mbgsinθ / (mb + mp)
a = 2.6m/s

α = a/r = (2.6m/s/s) / 0.5m = 5.22rads/s/s

CONCERNS:
We weren't given the final answer, so I'm not sure if my approach is correct. But I have a feeling that my Newton's 2nd Law (N2L) expression for the pulley is incorrect.
ΣFx = T = mpa
Since the pulley is rotating, would the acceleration in N2L be CENTRIPETAL ACCELERATION?
However, if this was the case, then the expression would turn into: ΣFx = T = mpv2/r

This introduces another unknown -- velocity! If this were the case, I don't know how to solve the problem.

Thank you!

2. Dec 4, 2015

### Orodruin

Staff Emeritus
No, you are not being asked for the acceleration of the pulley, it is likely considered fixed and rotating around its center. The pulley will therefore be assumed to have an additional constraining force at the center and so The force equation will give you nothing. You are asked for the angular acceleration, which is the rare of change of its angular velocity (how fast it rotates).

3. Dec 4, 2015

### vetgirl1990

I'm not quite sure what you mean by constraining force at the centre, and why the this would mean the force equation would give nothing. Would you be able to clarify?

Then, since I can't work with the block and pulley's force equations, I'm not quite sure where to start with finding the angular acceleration.
I'm thinking using Στ = Iα
Where the y-component of the block's gravity (mgsinθ) causes the torque (ie. the pulley to spin), and I = mr2

Στ = Iα
α = mbgsinθ / mpr2
Would this be a correct approach?

4. Dec 4, 2015

### Orodruin

Staff Emeritus
Correct, this is the torque equation for the pulley and it involves the angular acceleration which you are looking for.

It is not the block's gravity which causes the torque, it is the tension in the string. Because the block also accelerates, these two quantities will not be the same. The tension is an unknown which you will also have to solve for. Also note that the moment of inertia of a disk of mass m and radius r is not given by $I = mr^2$.

5. Dec 4, 2015

### vetgirl1990

Great, thank you -- this helps a lot! Instead, I'm using I = 1/2MR2. This is the moment of inertia for a disk.

One more thing... I know this is a pretty general question, but how do you properly identify the force that causes torque? Is it just finding the force that is directly connected to the rotating object's axis?
The reason why I thought the block's gravity caused torque was because it's the weight of the block that causes it to slide down the ramp, which therefore turns the pulley.

6. Dec 4, 2015

### haruspex

It is a common source of error to take an object as somehow experiencing a force to which it is not directly exposed. It is important to consider each object in quite a blinkered manner, only subject to forces that act on it directly.

7. Dec 6, 2015

### vetgirl1990

I'm taking a second look at this problem, and stuck on one part. I've figured out my net force and torque equations to be:

∑Fy = N - mgcosθ = 0 --> N = mgcosθ EQN 1
∑Fx = mgsinθ - T = ma --> T = mgsinθ - ma = mgsinθ - mrα --> T = m(gsinθ - rα) EQN 2
I still don't know what acceleration is, so I'm expressing it in terms of angular acceleration. a = rα

∑Torque = Iα = Trsinθ + (mgsinθ)rsinθ - Nrsinθ + (mgcosθ)rsinθ
All forces are perpendicular to the axis of rotation, so sinθ=1 and r cancels out... reducing the net equation to:

∑Torque = Iα = T + (mgsinθ) - N + (mgcosθ) EQN 3
Substituting EQN1 and EQN2 into EQN 3 to get rid of tension and normal force...
Iα = m(gsinθ - rα) + (mgsinθ) - (mgcosθ) + (mgcosθ)
(1/2MR2)α = m(gsinθ - rα) + (mgsinθ)

Plugging in all my known values (m=2kg, M=4kg, R=0.5m, θ=53°), I am still getting the wrong answer!
I played around a bit though, and noticed that if I did not express acceleration in terms of angular acceleration in my ∑Fx EQN 2, THEN I get the right answer.

Since the question states that the "block can slide down an incline, BUT connected to a pulley" does this mean that the block is actually not accelerating in the x direction, down the ramp?

8. Dec 6, 2015

### haruspex

Same mistake as before. Only include those forces that bear directly on the pulley.