Normal force of a bead moving around a horizontal ring

In summary: Yes. good job. Now, if you eliminate N between the two equations, you can solve for v as a function of t.
  • #1
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0
A bead of mass m is threaded on a metal hoop of radius R. There is a coefficient of kinetic friction µk between the bead and the hoop. It is given a push to start it sliding around the hoop with initial speed v0 . The hoop is located on the space station, so you can ignore gravity

Find the normal force exerted by the hoop on the bead as a function of its speed.

Not sure how to set this one up. I have a FBD with just normal force pointed towards the center of the ring, and the force of friction opposing the motion around the ring, and have confirmed that these are the only two forces acting on the bead.
 
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  • #2
And we're supposed to use a combination of these two equations.

Fr = −mv2
r

Ft = mat = m*dv
dt
 
  • #3
As you noted, the normal force is pointing in radially toward the center of the hoop. And you gave the equation for the radial force. So...

Chet
 
  • #4
I know with the radial force, I can easily rearrange the equation to solve for N in terms of v --> N = -mv^2 / R

But because of friction, v is changing with time, so shouldn't the answer be more complicated than that?
 
  • #5
laurenm02 said:
I know with the radial force, I can easily rearrange the equation to solve for N in terms of v --> N = -mv^2 / R

But because of friction, v is changing with time, so shouldn't the answer be more complicated than that?
No. They are only asking for the normal component of the acceleration. The friction affects only the tangential component of acceleration.

Chet
 
  • #6
Chestermiller said:
No. They are only asking for the normal component of the acceleration. The friction affects only the tangential component of acceleration.

Chet

Here's what I have:

There are two forces total acting on the bead--The normal force directed towards the center of the circle, and the force of kinetic friction opposing tangental velocity. Given the equation for radial force, I can solve for N = -mv^2 / R. Using friction, I can also solve for normal force, where -f = ma, and thus µk*N = ma and N = ma / µk and then N = (m / µk) * (dv/dt)
 
  • #7
laurenm02 said:
Here's what I have:

There are two forces total acting on the bead--The normal force directed towards the center of the circle, and the force of kinetic friction opposing tangental velocity. Given the equation for radial force, I can solve for N = -mv^2 / R. Using friction, I can also solve for normal force, where -f = ma, and thus µk*N = ma and N = ma / µk and then N = (m / µk) * (dv/dt)
Yes. good job. Now, if you eliminate N between the two equations, you can solve for v as a function of t.

Chet
 

1. What is the normal force of a bead moving around a horizontal ring?

The normal force of a bead moving around a horizontal ring is the force exerted by the ring on the bead perpendicular to the surface of the ring. It acts as a support force to keep the bead from falling off the ring.

2. How is the normal force affected by the speed of the bead?

The normal force is not affected by the speed of the bead. It remains constant as long as the bead is moving at a constant speed around the ring.

3. Does the normal force change if the ring is tilted?

Yes, the normal force will change if the ring is tilted. As the angle of the ring increases, the normal force will decrease because the component of the gravitational force acting on the bead parallel to the surface of the ring will increase.

4. Can the normal force be greater than the weight of the bead?

Yes, it is possible for the normal force to be greater than the weight of the bead. This can happen if the ring is tilted at a steep angle, causing the normal force to increase in order to balance the increased gravitational force acting on the bead.

5. How does the normal force affect the motion of the bead?

The normal force does not directly affect the motion of the bead, but it is necessary for the bead to maintain circular motion around the ring. Without the normal force, the bead would not have a support force and would fall off the ring.

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