Torque and rigid body rotation


by transparent
Tags: body, rigid, rotation, torque
transparent
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#1
May23-13, 02:26 PM
P: 16
Hello. My first post here. I'm having trouble with the basics of rigid body rotation. I have a few questions (my apologies if they are too childish; I'm very new to this):

1) Is torque (and other angular parameters like angular velocity, angular acceleration etc.) defined about a point or an axis? If about an axis, then can we choose any point on the axis as origin when calculating r x f?

2) Is a couple always required to produce rotation? If so, then how can torque be produced with just one force? If we have a rod (no gravity, no friction) and we apply force at one of its end, then will it start rotating? The force will produce torque about all the points on the rod except the end, but there will be no couple.

3) In rotational equilibrium, does the net torque about every point/axis have to be zero? Is this even possible? I mean, any force will produce torque about some point.

Thanks in advance to anyone who can help me out.

Edit: I think I've posted in the wrong section. How do I move/delete this thread?
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rock.freak667
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#2
May24-13, 09:09 PM
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Quote Quote by transparent View Post
Hello. My first post here. I'm having trouble with the basics of rigid body rotation. I have a few questions (my apologies if they are too childish; I'm very new to this):

1) Is torque (and other angular parameters like angular velocity, angular acceleration etc.) defined about a point or an axis? If about an axis, then can we choose any point on the axis as origin when calculating r x f?
The torque given by r x f will produce a vector which is both perpendicular to r and f. So if you are to calculate the torque about a point due to a force F. Then you just need to get the r vector from the point to the force and then apply the cross-product. I am not sure if I answered your question here.

Quote Quote by transparent View Post
2) Is a couple always required to produce rotation? If so, then how can torque be produced with just one force? If we have a rod (no gravity, no friction) and we apply force at one of its end, then will it start rotating? The force will produce torque about all the points on the rod except the end, but there will be no couple.
A couple would consist of two identical forces in opposite directions. So if you consider a rigid bar fixed at one end (like a hinge so it can rotate). If you apply a force about the free end (not fixed end), the bar will rotate and the force would not be due to a couple.

Quote Quote by transparent View Post
3) In rotational equilibrium, does the net torque about every point/axis have to be zero? Is this even possible? I mean, any force will produce torque about some point.
Yes a condition of equilibrium would be the torque about any point would be zero. Based on the problem, if you apply your equilibrium equations (ƩFx=Fy=Fz=0), you will get reaction forces, which when if you calculate the torque about a single point, the reaction forces will produce torques which can cause the overall net torque to be equal to zero.
Doc Al
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#3
May26-13, 06:46 AM
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Quote Quote by transparent View Post
1) Is torque (and other angular parameters like angular velocity, angular acceleration etc.) defined about a point or an axis? If about an axis, then can we choose any point on the axis as origin when calculating r x f?
In general, r x f is the torque about a point. If you just want the torque about some axis, you'd take the component of that torque along that axis. You can choose any point along the axis as your origin--the component of the torque parallel to the axis will be the same.

transparent
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#4
May27-13, 01:27 AM
P: 16

Torque and rigid body rotation


Quote Quote by rock.freak667 View Post
A couple would consist of two identical forces in opposite directions. So if you consider a rigid bar fixed at one end (like a hinge so it can rotate). If you apply a force about the free end (not fixed end), the bar will rotate and the force would not be due to a couple.
What if the rigid bar is not fixed at one end, as in my question? Will the torque still make it rotate?

Quote Quote by Doc Al View Post
In general, r x f is the torque about a point. If you just want the torque about some axis, you'd take the component of that torque along that axis. You can choose any point along the axis as your origin--the component of the torque parallel to the axis will be the same.
Thanks.
butan1ol
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#5
May27-13, 02:20 AM
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Quote Quote by transparent View Post
What if the rigid bar is not fixed at one end, as in my question? Will the torque still make it rotate?
When the rigid bar is not fixed at one end, the torque still make it rotate, and the single force will cause translational acceleration on the bar.

While for couple, the pair of forces gives no resultant forces. It is just rotation without translation.
transparent
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#6
May28-13, 02:04 AM
P: 16
Quote Quote by butan1ol View Post
When the rigid bar is not fixed at one end, the torque still make it rotate, and the single force will cause translational acceleration on the bar.

While for couple, the pair of forces gives no resultant forces. It is just rotation without translation.
Thanks.


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