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