# Normal force of a bead moving around a horizontal ring

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1. Jan 30, 2015

### laurenm02

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.

2. Jan 30, 2015

### laurenm02

And we're supposed to use a combination of these two equations.

Fr = −mv2
r

Ft = mat = m*dv
dt

3. Jan 30, 2015

### Staff: Mentor

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. Jan 30, 2015

### laurenm02

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. Jan 30, 2015

### Staff: Mentor

No. They are only asking for the normal component of the acceleration. The friction affects only the tangential component of acceleration.

Chet

6. Jan 30, 2015

### laurenm02

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. Jan 31, 2015

### Staff: Mentor

Yes. good job. Now, if you eliminate N between the two equations, you can solve for v as a function of t.

Chet