# Non-Uniform circular motion with friction

1. Nov 4, 2007

### Knissp

1. The problem statement, all variables and given/known data
A small block of mass m slides on a horizontal frictionless surface as it travels around the inside of a hoop of radius R. The coefficient of friction between the block and the wall is mu; therefore, the speed v of the block decreases. In terms of m, R. mu, and v, find expressions for each of the following.
b. The block's tangential acceleration dv/dt
c. The time required to reduce the speed of the block from an initial value v0 to vo/3
(A diagram is given here on page 4: www.swcp.com/~gants/calendars/ap%20physics/sept%20docs/(A)%20Newton's%20LawsC.doc)

2. Relevant equations
dv/dt = aT = tangential acceleration
aA = angular acceleration
aC = centripetal acceleration
v/r=w (angular velocity)

3. The attempt at a solution
First I tried to use aT = r * aA = r * d^2 (theta) / dt^2 but I don't think that's right because none of those values are given.
Then I tried to do
Fnet = m * (aC^2 + aT^2)^(1/2) = ((m*v^2/r)^2+(m*aT - mu*m*v^2/r))^(1/2)
but I think that's wrong because I'm taking taking the forces and finding the vector sum of them and I don't know if that's allowed like I can do with the (aC^2 + aT^2)^(1/2) to get acceleration.

Perhaps would it help for me to know that dv/dt = dv/dx * dx/dt = v*dv/dx?? I'm kind of lost, any help would be greatly appreciated.

Last edited: Nov 4, 2007
2. Nov 4, 2007

### Staff: Mentor

The only tangential force is friction. What does that equal? (Hint: What's the normal force?)

3. Nov 4, 2007

### Knissp

Is this correct?

F = m * dv/dt = mu*m*v^2/r

dv/dt = mu*v^2/r

4. Nov 4, 2007

### Staff: Mentor

Looks good to me.

5. Nov 4, 2007

### Knissp

Yay. Thank you so much.

6. Nov 4, 2007

### Knissp

Wait, tangential acceleration is supposed to be negative, right? So is it supposed to be -mu*v^2/r or is something done wrong?

And just to make sure, for part c where it asks about the time required to reduce speed from v0 to 1/3*v0, is it:
dv/dt = a = -mu*v^2/r

2/3*v / delta(t) = -mu*v^2/r

(-2*r)/(3*mu*v) = delta(t)

But that doesn't make sense because the change in time cannot be negative.

-EDIT: never mind about the time being negative part. haha i know the change in velocity has to be -2/3 because its 1/3 v - 1 v. and the negative of the negative 2/3 makes it positive.
but can you still check everything else?

Last edited: Nov 4, 2007
7. Nov 4, 2007

### Staff: Mentor

Right. It should be negative.

That should be:
dv/dt = -mu*v^2/r

This is not correct. Don't assume the acceleration is constant (the velocity is not). Integrate! ($dv/dt \neq \Delta v / \Delta t$)

8. Nov 4, 2007

### Knissp

Is this right?

dv/dt = a = -mu*v^2/r

integral(dv/v^2) = integral (-mu/r dt)

-1/v = -mu/r * delta(t)

delta (t) = r/(v*mu)

And isn't acceleration constant? I know velocity is not.

9. Nov 4, 2007

### Staff: Mentor

Good.

Evaluate the integral properly:
$$\int_{v_0}^{v_0/3}(1/v^2) dv = -\int\mu/r dt$$

How can it be constant if it depends on velocity?

Last edited: Nov 4, 2007