How Does a Spool of Thread Accelerate When Subjected to Force?

[frillybob]

Homework Statement

A spool of thread consists of a cylinder of radius R1 = 6.6 cm with end caps of radius R2 = 9.3 cm as depicted in the end view shown in the figure below. The mass of the spool, including the thread, is m = 290 g. The spool is placed on a rough, horizontal surface so that it rolls without slipping when a force

= 0.690 N acting to the right is applied to the free end of the thread. For the moment of inertia treat the spool as being a solid cylinder of radius R1, as the extended edges are thin and therefore light.

(a) What is the acceleration of the spool? Take positive to be to the right.

http://www.webassign.net/serpse8/10-p-085.gif

Homework Equations

Solid Cylinder I = 1/2(M)(r)^2

The Attempt at a Solution

[/B]
I need some help with this problem. I don't think my attempt is even correct. But my thought is since T = I*A. Then I just need to solve for A. A = T*I. The moment of inertia is what is stumping me. I'm not sure if I'm doing this correctly. I = 1/2(.3)(.066) = 0.0099. This does not solve the problem adequately. Thanks of the help!

*Where T is force
A is angular acceleration
I is moment of inertia.

 

[ehild]

What is the question?

 

[frillybob]

What is the question?

Updated in main post. Sorry about that!

 

[ehild]

Homework Statement

A spool of thread consists of a cylinder of radius R1 = 6.6 cm with end caps of radius R2 = 9.3 cm as depicted in the end view shown in the figure below. The mass of the spool, including the thread, is m = 290 g. The spool is placed on a rough, horizontal surface so that it rolls without slipping when a force

= 0.690 N acting to the right is applied to the free end of the thread. For the moment of inertia treat the spool as being a solid cylinder of radius R1, as the extended edges are thin and therefore light.

(a) What is the acceleration of the spool? Take positive to be to the right.

http://www.webassign.net/serpse8/10-p-085.gif

Homework Equations

Solid Cylinder I = 1/2(M)(r)^2

The Attempt at a Solution

[/B]
I need some help with this problem. I don't think my attempt is even correct. But my thought is since T = I*A. Then I just need to solve for A. a = T*I. The moment of inertia is what is stumping me. I'm not sure if I'm doing this correctly. I = 1/2(.3)(.066) = 0.0099. This does not solve the problem adequately. Thanks of the help!

What are A and a? T is the force in the problem. Better to denote it F.

 

[frillybob]

What are A and a? T is the force in the problem. Better to denote it F.

Updated in main post. I left it as T because that is what the problem defines it by.

 

[rcgldr]

The horizontal surface applies a force to the left at the edges of the end caps as the spool accelerates. This torque is partially opposed by the thread's force to the right at the edge of the inner part of the spool.

 

[ehild]

The horizontal surface applies a force to the left at the edges of the end caps as the spool accelerates. This torque is partially opposed by the thread's force to the right at the edge of the inner part of the spool.

The torque is defined with respect a point or axis. What is your point of reference?

 

[frillybob]

The torque is defined with respect a point or axis. What is your point of reference?

The question says to take positive to be the right. It is moving to the right as well.

 

[ehild]

How do you calculate torque?

 

[frillybob]

How do you calculate torque?

T=I*A

Where T is torque (Given)
I is moment of inertia (Not given)
A is angular Acceleration (Need to find)

 

[ehild]

T=I*A

Where T is torque (Given)
I is moment of inertia (Not given)
A is angular Acceleration (Need to find)

Yes, but how is the torque of a force defined?

 

[frillybob]

Yes, but how is the torque of a force defined?

Torque is a measure of how much a force acting on an object causes that object to rotate.

https://www.physics.uoguelph.ca/tutorials/torque/Q.torque.intro.html

 

[ehild]

Torque is a measure of how much a force acting on an object causes that object to rotate.

https://www.physics.uoguelph.ca/tutorials/torque/Q.torque.intro.html

Well, how much is the torque of the applied force in this case?

 

[frillybob]

Well, how much is the torque of the applied force in this case?

I assume all of it because it says it doesn't slip.

 

[ehild]

Answer my question, please.

 

[frillybob]

Answer my question, please.

0.690N?

Not trying to irritate you or anything. It's my first time on this board.

 

[ehild]

0.690N?

Not trying to irritate you or anything. It's my first time on this board.

The torque is not force.

 

[frillybob]

The torque is not force.

F = M*A
2.94N = 0.3Kg*9.8m/s/s?

 

[SteamKing]

Homework Statement

A spool of thread consists of a cylinder of radius R1 = 6.6 cm with end caps of radius R2 = 9.3 cm as depicted in the end view shown in the figure below. The mass of the spool, including the thread, is m = 290 g. The spool is placed on a rough, horizontal surface so that it rolls without slipping when a force

= 0.690 N acting to the right is applied to the free end of the thread. For the moment of inertia treat the spool as being a solid cylinder of radius R1, as the extended edges are thin and therefore light.

(a) What is the acceleration of the spool? Take positive to be to the right.

http://www.webassign.net/serpse8/10-p-085.gif

Homework Equations

Solid Cylinder I = 1/2(M)(r)^2

The Attempt at a Solution

[/B]
I need some help with this problem. I don't think my attempt is even correct. But my thought is since T = I*A. Then I just need to solve for A. A = T*I. The moment of inertia is what is stumping me. I'm not sure if I'm doing this correctly. I = 1/2(.3)(.066) = 0.0099. This does not solve the problem adequately. Thanks of the help!

*Where T is force
A is angular acceleration
I is moment of inertia.

Check your arithmetic. You calculated an incorrect value for I.

 

[ehild]

F = M*A
2.94N = 0.3Kg*9.8m/s/s?

I asked the torque.
I leave you now, I have to sleep. Read your notes or handbook or the page you cited about the torque. You might remember that you get it multiplying the force with its lever arm.
http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html

 

[ehild]

T=I*A

Where T is torque (Given)
I is moment of inertia (Not given)
A is angular Acceleration (Need to find)

The problem says that T is force, 0.690 N.
If it was torque, about what point is it?

 

[rcgldr]

The horizontal surface applies a force to the left at the edges of the end caps as the spool accelerates. This torque is partially opposed by the thread's force to the right at the edge of the inner part of the spool.

The torque is defined with respect a point or axis. What is your point of reference?

I'm considering the axis of the spool as the center of rotation for the net torque. If viewing this as an instant in time, then the point of contact with the surface could also be used, in which case the ground force doesn't matter, so the only torque is related to the tension in the string times the distance above the surface, but the moment of inertia about the point of contact becomes more complicated. The problem statement defines the moment of inertia about the axis of the spool, which is why I chose that as the reference point for the torques.

Back to the original question, note that the linear acceleration equals the net linear force divided by mass. The net linear force equals the tension to the right minus the surface force to the left. The angular acceleration equals the linear acceleration divided by the radius of the end caps.

 

[haruspex]

Torque is a measure of how much a force acting on an object causes that object to rotate.

https://www.physics.uoguelph.ca/tutorials/torque/Q.torque.intro.html

That's a poor definition, discard it. Torque is meaningful given any force and any reference point or reference axis. There does not need to be a body present that can or will rotate about that point.
You might find https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/ helpful.

 

[frillybob]

I'm considering the axis of the spool as the center of rotation for the net torque. If viewing this as an instant in time, then the point of contact with the surface could also be used, in which case the ground force doesn't matter, so the only torque is related to the tension in the string times the distance above the surface, but the moment of inertia about the point of contact becomes more complicated. The problem statement defines the moment of inertia about the axis of the spool, which is why I chose that as the reference point for the torques.

Back to the original question, note that the linear acceleration equals the net linear force divided by mass. The net linear force equals the tension to the right minus the surface force to the left. The angular acceleration equals the linear acceleration divided by the radius of the end caps.

Thanks for that. Could you happen to explain your reasoning to the second paragraph a bit more? Could you also explain how to find it. (You can use different numbers than mine, I work much better with numbers)

 

[haruspex]

I work much better with numbers)

I very strongly encourage you to develop the habit of working as far as possible without numbers. It has many advantages, and you will get better at it with practice.

With regard to the present problem, I consider it simpler to take moments about the point of contact with the ground. You just need to apply the parallel axis theorem, which I assume you have met. But go with rcgldr for now and we can look at that for comparison later. Anyway, I see this is part a) only. Depending on what the other parts ask, there might be no advantage.

 

[rcgldr]

Since the originator hasn't posted in a while, this is what I get for the friction force (with direction to left), as a function of tension (with direction to right), r1, and r2, and angular inertia = 1/2 m r1^2. I left out the steps leading up to this in case the originator decides to continue.

f = t(r1 + 2) /(r1 + 2 (r2/r1))
linear acceleration = (t - f) / m

update - correction:

f = t (r1 r2 + (1/2 r1^2)) / (1/2 r1^2 + r2^2)

linear acceleration = (t - f) / m = (t r2 ( r2 - r1)) / m (1/2 r1^2 + r2^2)

 

[TSny]

f = t(r1 + 2) /(r1 + 2 r2/r1)

Do a dimensional analysis of the quantities in parentheses.

 

[haruspex]

Since the originator hasn't posted in a while, this is what I get for the friction force (with direction to left), as a function of tension (with direction to right), r1, and r2, and angular inertia = 1/2 m r1^2. I left out the steps leading up to this in case the originator decides to continue.

f = t(r1 + 2) /(r1 + 2 r2/r1)
linear acceleration = (t - f) / m

As TSny points out, you must have some typos in there.
Anyway, taking moments about point of contact produces immediately $\alpha=\frac{T(r_2-r_1)}{m(\frac 12r_1^2+r_2^2)}$

 

[rcgldr]

As TSny points out, you must have some typos in there.
Anyway, taking moments about point of contact produces immediately $\alpha=\frac{T(r_2-r_1)}{m(\frac 12r_1^2+r_2^2)}$

I left out an r2 term. Will note a correction to my previous post.

I ended up with

f = t (r1 r2 + (1/2 r1^2)) / (1/2 r1^2 + r2^2)

linear acceleration = (t - f) / m = (t r2 ( r2 - r1)) / (m (1/2 r1^2 + r2^2))

angular acceleration = linear acceleration / r2 = t (r2 - r1) / (m (1/2 r1^2 + r2^2))

As pointed out by haruspex, parallel axis theorem makes this much easier, but I don't know if that was covered in the originator's class yet.