Acceleration without resultant force (rotating body)

  • Thread starter pinsky
  • Start date
  • #1
96
0

Main Question or Discussion Point

Hello there.

I have some difficulty in understanding the condition for a rigid body standing still (or rotating).

If we have a body, lets say a yoyo which is somehow pierced through the center and attached in a way that it's center of mass can't move, but the yoyo can rotate if we apply a torque.

If we pull the yoyo rope, it will exert a torque which will make the yoyo start rotating, but there will be a force of same size but opposite direction acting on the center of rotation. This force wont exert a torque (since it acts on the rotation axis) but it will prevent translation movement of the center of mass since all forces applied on the body are zero.

Now let's observe this image:

[PLAIN]http://img524.imageshack.us/img524/4026/kuglananagibu.gif [Broken]

And lets suppose

[tex]k mg Cos{\alpha}-mg Sin{\alpha}=0[/tex]

The resultant torque is:

[tex]k mg Cos{\alpha} R=J \alpha[/tex]

(R is the radius and alpha in this case is the angular acceleration)

So in the end we have a rolling body which has a translational acceleration of the center of mass, without any resultant force acting on the body.

I know that can't be, i just don't know what i'm I doing wrong.

Please help
 
Last edited by a moderator:

Answers and Replies

  • #2
AlephZero
Science Advisor
Homework Helper
6,994
291
Hello there.

I have some difficulty in understanding the condition for a rigid body standing still (or rotating).

If we have a body, lets say a yoyo which is somehow pierced through the center and attached in a way that it's center of mass can't move, but the yoyo can rotate if we apply a torque.

If we pull the yoyo rope, it will exert a torque which will make the yoyo start rotating, but there will be a force of same size but opposite direction acting on the center of rotation. This force wont exert a torque (since it acts on the rotation axis) but it will prevent translation movement of the center of mass since all forces applied on the body are zero.
What you said there is correct.

And lets suppose

[tex]k mg Cos{\alpha}-mg Sin{\alpha}=0[/tex]
Your mistake is assuming that is true. It isn't.

Draw a free body diagram showing all the forces on the disk. Resolve the weight of the disk into the components normal and parallel to the slope. There are also normal and tangential forces where the disk touches the plane. Call them [tex]F_n[/tex] and [tex]F_t[/tex].

There is no acceleration of the disk normal to the plane, so the two normal forces must be equal and opposite.

If you are assuming the disk rolls without slipping, then the linear velocity must be related to the angular velocity by
[tex]v = \omega r[/tex]
The linear and angular accelerations are related by the same equation.

So you have an equation for the linear acceleration down the slope

[tex]mg \cos\alpha - F_t = ma = mr \dot\omega[/tex]

And an equation for the angular accleration of the disk, taking moments about its center:

[tex]r F_t = J \dot\omega[/tex]

You can solve those two equations in the two variables [tex]F_t[/tex] and [tex]\dot\omega[/tex]. You don't need to make any assumption about [tex]F_t[/tex].
 
  • #3
Doc Al
Mentor
44,882
1,129
And lets suppose

[tex]k mg Cos{\alpha}-mg Sin{\alpha}=0[/tex]
As AlephZero points out, that isn't true.

Furthermore, what's k? The coefficient of static friction? Realize that the frictional force does not necessarily equal kN.
 
  • #4
96
0
Thanks for your reply.

But

[tex]
k mg Cos{\alpha}-mg Sin{\alpha}=0
[/tex]

is the situation i want to observe. Or is it physically imposable for it to occur? An if it is, why is that?
 
  • #5
Doc Al
Mentor
44,882
1,129
But

[tex]
k mg Cos{\alpha}-mg Sin{\alpha}=0
[/tex]

is the situation i want to observe. Or is it physically imposable for it to occur? An if it is, why is that?
How do you think you'd arrange for this condition to be met?
 
  • #6
96
0
[tex]k = tg{\alpha}[/tex] ?
 
  • #7
Doc Al
Mentor
44,882
1,129
[tex]k = tg{\alpha}[/tex] ?
Not sure what you mean. Are you thinking that if you choose the right value for the coefficient of static friction then your condition would be met? If so, realize that the friction required for a ball rolling without slipping down an incline does not depend on the coefficient of friction (as long as that coefficient is above some minimum value).
 
  • #8
AlephZero
Science Advisor
Homework Helper
6,994
291
The only way you could observe that situation is put some glue on the plane. Then you would be right, in the sense that the disk won't roll down the plane :smile:

When the disk is rolling without slipping, there is static friction between the disk and the plane, not dynamic friction.

For dynamic friction, when there is sliding, and the coefficient of friction is k, then you know the friction force is equal to k times the normal force.

For static friction, with no sliding, all you know is that the friction force is less than or equal to k times the normal force. The actual magnitude of the friction force is whatever it needs to be, to satisfy Newton's laws of motion.

For example if you have a box weighing 100N on a horizontal plane and the coefficienct of static friction is 0.5, then if you push the box horizontally with a force of 10N, the box doesn't move and the friction force is 10N. If you push with 50N, the box still doesn't move and the friction force is 50N

For the wheel problem, the friction force is just the right size to make the wheel roll without slipping. You can't do anything to change that.

If the static friction force is too small to make the wheel roll without slipping (either the static friction coefficient is small, or the plane is at steep angle) then there will be a combination of sliding and rolling, depending on the size of the dynamic friction coefficient. (And as I said earlier, the dynamic friction force is a known value, if you know the normal force and the friction coefficient).
 

Related Threads for: Acceleration without resultant force (rotating body)

Replies
6
Views
4K
Replies
4
Views
1K
  • Last Post
Replies
1
Views
939
Replies
8
Views
5K
Replies
3
Views
1K
Replies
6
Views
449
Top