Finding angular acceleration of pulley attached to a mass

In summary, the block's weight (mg) is balanced by the normal force (N) in the y-direction (perpendicular to the incline). In the x-direction (parallel to the incline), the tension in the rope (T) is equal to the block's weight (mg) multiplied by the sine of the incline angle (θ) minus the product of the block's mass (m) and its angular acceleration (α). The torque equation is found by summing the torques exerted by each of these forces on the pulley, which is equal to the moment of inertia (I) multiplied by the angular acceleration (α).
  • #1
vetgirl1990
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Homework Statement


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?

Homework Equations


a = rα

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!
 
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  • #2
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
Orodruin said:
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).

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
vetgirl1990 said:
I'm thinking using Στ = Iα
Correct, this is the torque equation for the pulley and it involves the angular acceleration which you are looking for.

vetgirl1990 said:
Where the y-component of the block's gravity (mgsinθ) causes the torque (ie. the pulley to spin), and I = mr2

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
Orodruin said:
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##.

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
vetgirl1990 said:
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.
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.
 
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  • #7
haruspex said:
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.

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
vetgirl1990 said:
∑Torque = Iα = T + (mgsinθ) - N + (mgcosθ) EQN 3
Same mistake as before. Only include those forces that bear directly on the pulley.
 
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1. How do you calculate the angular acceleration of a pulley attached to a mass?

To calculate the angular acceleration of a pulley attached to a mass, you will need to use the formula α = (τ - Iα)/I, where α is the angular acceleration, τ is the torque, and I is the moment of inertia of the pulley. You will also need to know the mass and radius of the pulley.

2. What is the relationship between angular acceleration and linear acceleration?

Angular acceleration and linear acceleration are related through the formula α = a/r, where α is the angular acceleration, a is the linear acceleration, and r is the radius of the pulley. This means that for a given linear acceleration, the angular acceleration will increase as the radius decreases.

3. How does the mass attached to the pulley affect the angular acceleration?

The mass attached to the pulley does not directly affect the angular acceleration. However, it does affect the torque applied to the pulley, which in turn affects the angular acceleration. A larger mass will result in a larger torque and therefore a larger angular acceleration.

4. Can the angular acceleration of a pulley attached to a mass be negative?

Yes, the angular acceleration of a pulley attached to a mass can be negative. A negative angular acceleration means that the pulley is slowing down in its rotation.

5. What factors can affect the accuracy of calculating the angular acceleration of a pulley attached to a mass?

There are a few factors that can affect the accuracy of calculating the angular acceleration of a pulley attached to a mass. These include friction, the accuracy of the measurements of the pulley's mass and radius, and any external forces acting on the pulley.

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