# Angular Acceleration and torque problem

#### BrainMan

1. Homework Statement
A 12 kg mass is attached to a cord that is wrapped around a wheel of radius r = 10 cm (Fig. 8.20). The acceleration of the mass down the friction-less incline is 2.0 m/s^2/. Assuming the axle of the wheel to be friction-less, determine (a) the tension in the cord, (b) the moment of inertia of the wheel , and (c) the angular speed of the wheel 2 s after it begins rotating, starting from rest.

2. Homework Equations
F = ma

3. The Attempt at a Solution
I found the tangential force by using F = ma and found it to be 24. I am not sure how to continue to find the angular acceleration and moment of inertia because the question doesn't give the mass of the wheel.

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#### Nathanael

Homework Helper
You're not given the mass of the wheel, but you are given the acceleration of the object.

If you were given the mass of the wheel instead of the acceleration of the object, would you be able to figure out the acceleration?

If so, then you should be able to figure out the mass given the acceleration.

#### BrainMan

You're not given the mass of the wheel, but you are given the acceleration of the object.

If you were given the mass of the wheel instead of the acceleration of the object, would you be able to figure out the acceleration?

If so, then you should be able to figure out the mass given the acceleration.
OK so I need to find the acceleration of the block down the hill and that will be the acceleration on the wheel? If so how do I do that?

#### Nathanael

Homework Helper
OK so I need to find the acceleration of the block down the hill and that will be the acceleration on the wheel? If so how do I do that?
You don't need to find the acceleration of the block down the hill, because the problem has already told you that it is 2 m/s^2

The linear acceleration of the block is equal to the linear acceleration of the edge of the wheel.
(This assumes the rope doesn't slip on the pulley, which you need to assume for the problem to be solvable.)

This gives you the equation:
a $=\alpha r$
Where "a" is the linear acceleration (=2 m/s^2) and "$\alpha$" is the angular acceleration of the wheel.

#### BrainMan

You don't need to find the acceleration of the block down the hill, because the problem has already told you that it is 2 m/s^2

The linear acceleration of the block is equal to the linear acceleration of the edge of the wheel.
(This assumes the rope doesn't slip on the pulley, which you need to assume for the problem to be solvable.)

This gives you the equation:
a $=\alpha r$
Where "a" is the linear acceleration (=2 m/s^2) and "$\alpha$" is the angular acceleration of the wheel.
OK I did that and got the last part of the problem right. However, I am still having trouble with the second part of the problem. So

Ft = ma
Ft = 12(2)
Ft = 24
T = Ft*r
T = 24 * .1
T = 2.4
T = I*a
T = I * 20
2.4 = I *20
I = .12

The correct answer is 0.234 kg * m2

#### haruspex

Homework Helper
Gold Member
2018 Award
OK I did that and got the last part of the problem right. However, I am still having trouble with the second part of the problem. So

Ft = ma
Ft = 12(2)
Ft = 24
T = Ft*r
T = 24 * .1
T = 2.4
T = I*a
T = I * 20
2.4 = I *20
I = .12

The correct answer is 0.234 kg * m2
Ft is the net downslope force on the block, right? But you have equated that to the tension, which is an applied upslope force, and neglected gravity.

#### BrainMan

Ft is the net downslope force on the block, right? But you have equated that to the tension, which is an applied upslope force, and neglected gravity.
Why do I have to use gravity it already tells me that the block is accelerating down the slop at 2 m/s^2?

#### Nathanael

Homework Helper
Why do I have to use gravity it already tells me that the block is accelerating down the slop at 2 m/s^2?
Because the two forces on the block are gravity and tension (in opposite directions)

You don't know the tension, but you do know the gravity and you know the net acceleration (and thererfore net force) so you can use that to find the tension

#### BrainMan

Because the two forces on the block are gravity and tension (in opposite directions)

You don't know the tension, but you do know the gravity and you know the net acceleration (and thererfore net force) so you can use that to find the tension
So do I do
Fg + A = T?

Homework Helper
Fg - T = M*a

Fg - Ma = T

#### Nathanael

Homework Helper
How did you get these equations?
Well the second one is just a rearrangement of the first one.

I forgot to mention: "Fg" is not mg, it's the just the component of mg down the slope.

I got the equation because "Fg" (the force of gravity down the slope) is pulling the object down the slope, while the tension "T" is pulling the object up the slope.

The net acceleration is "a" down the slope so the net force is "m*a" (directed down the slope)

Since there is no other forces acting (the normal force is cancelled by a component of gravity) those two forces ("Fg" and "T") must add to give "ma" (They are in opposite directions, so that's where the negative sign comes from)

This wasn't worded very clearly, but I hope you understand what I was trying to say

#### rcgldr

Homework Helper
How did you get these equations?
The acceleration of the block is the net force on the block divided by the mass of the block.

The component of force due to gravity on a frictionless incline = m g sin(θ)
The opposing force is the tension in the rope = T.

The net force on the block in the direction of slope of the incline = m g sin(θ) - T

The acceleration of the block = (m g sin(θ) - T) / m = g sin(θ) - T / m = g sin(37°) - T / 12kg = 2 m / s^2, use this to solve for T. The unknown here is what value to use for g, 9.80665, 9.8, 9.81, ... ?

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#### BrainMan

Well the second one is just a rearrangement of the first one.

I forgot to mention: "Fg" is not mg, it's the just the component of mg down the slope.

I got the equation because "Fg" (the force of gravity down the slope) is pulling the object down the slope, while the tension "T" is pulling the object up the slope.

The net acceleration is "a" down the slope so the net force is "m*a" (directed down the slope)

Since there is no other forces acting (the normal force is cancelled by a component of gravity) those two forces ("Fg" and "T") must add to give "ma" (They are in opposite directions, so that's where the negative sign comes from)

This wasn't worded very clearly, but I hope you understand what I was trying to say
So gravity is not causing the 2 m/s^2 acceleration. I understand how you got the variables but I am still confused on the relationships between them. I know that the force of gravity and the force of the acceleration are both in the same direction and are resisted by the tension. However, the formula is not Fg + ma = T. Why is this?

#### Nathanael

Homework Helper
So gravity is not causing the 2 m/s^2 acceleration. I understand how you got the variables but I am still confused on the relationships between them. I know that the force of gravity and the force of the acceleration are both in the same direction and are resisted by the tension. However, the formula is not Fg + ma = T. Why is this?
Because the "force of acceleration" is not really a force in the system.

You know (from Fnet=ma) that the net force on the block has to be ma, but ma is not really a force acting on the block, it's just the result of combining the other forces (tension and gravity)

#### rcgldr

Homework Helper
So gravity is not causing the 2 m/s^2 acceleration.
Take a look at post #14, which explains the math in more detail.

#### BrainMan

Take a look at post #14, which explains the math in more detail.
OK so I did that and am still getting the question wrong so
(g sin 37 -T)/12 = 2
g sin 37 -T = 24
5.898 - T = 24
T = -18.102
The book says the correct answer is 46.8 N

#### Nathanael

Homework Helper
It acidentally double posted this... Is there a way to delete posts?

@rcgldr, I don't see a delete button. (I've seen it, and used it, before, but it's now nowhere to be found)

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#### Nathanael

Homework Helper
it should be:

g sin(37°) - T/12 = 24

(The "g sin(37)" is already divided by 12, because it was originally "mg sin(37)" so you don't need to divide it by 12 again)

#### rcgldr

Homework Helper
Continuing from post #14, as mentioned g sin(37°) was already divided by the mass of 12 kg:

g sin(37°) - T / 12kg = 2 m / s^2

(9.80665) (0.6018) - T / 12 = 2
T/12 = (9.80665) (0.6018) - 2 = 5.9018 - 2 = 3.9018
T = (12) (3.9018) = 46.82 N

@Nathanael - yes you can delete a post. Select edit, and you'll see a "Delete" button.

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#### dean barry

Heres my take on this problem :

g assumed at 9.81 (m/s)/s

The force down the slope created by the mass of the block = m * g * ( sine ( 37 ° ) )
= 70.8457 N

The force required to accelerate the 12 kg mass @ 2.0 ( m / s ) / s
= m * a
= 24 N

The remainder must be driving the wheel = 70.8457 - 24 = 46.8457 N
(this is the tension on the string)

The torque on the wheel = 46.8457 * 0.1
= 4.68457 N-m

The angular acceleration = 2.0 / 0.1
= 20.0 ( rad / sec ) / sec

The mass moment of inertia :
= torque / angular acceleration
= 4.68457 / 20.0
= 0.2342 kg-m²

The angular rotation rate after 2 seconds :
= angular acceleration * time
= 20.0 * 2.0

#### Nathanael

Homework Helper
Dean Barry, you're supposed to let the OP solve the problem.

Guidelines for students and helpers said:
9. Helpers: don't provide the entire solution

"Angular Acceleration and torque problem"

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