Homework Help: Derivation of Tangential Acceleration in Non-Uniform Circular Motion

1. Nov 6, 2008

drummerteenX

1. The problem statement, all variables and given/known data
A small block of mass "m" slides on a frictionless horizontal surface as it travels inside of a hoop of radius "R". The coefficient of friction between the block and the wall is "u", therefore, the speed of the block decreases. In terms of "m", "R", "u", and "v" (the block's velocity), find expressions for the following.
1. The frictional force on the block: (I believe it is f = (v^(2) * m / R)

2. The block's tangential acceleration (dv/dt). I really need this one.

3. Use the equation from #2 to find the time required to reduce the speed of the block to one-third of its original velocity from its original velocity.

2. Relevant equations

N/A

3. The attempt at a solution

#1 should be "f = (v^(2) * m / R"

#2 All I have so far is #1 solved for "v". "v = sqrt(f * R / m)"

#3 Once I have #2, this will be cake.

2. Nov 6, 2008

LowlyPion

Welcome to PF.

In one you have ignored the coefficient of friction. Your expression if for the radial force. Frictional force is Normal force * coefficient of friction.

From F = m*a , knowing 1, you can solve immediately for 2.

Since 3 is a piece of cake ... you're all set.

3. Mar 28, 2010

tskillet01

Ok, so I solved for a) and b), but c) is not "cake" for me. Can you help me out for where to start solving it?

4. Mar 28, 2010

AtticusFinch

Ok so this is what I get for

a) -u*v^(2)*m/R = F

b)dv/dt = -u*v^(2)/R

If you accept those answers then this is how I would solve c)

dv = (-u*v^(2)/R)dt

dv*R/(-u*v^2)= dt

Integrate both sides. Left integral is from v to 1/3v. Right integral is from 0 to t.

After the integration you get this:

R/(u*v) = t left side still needs to be evaluated on limits of integration so...

R/(u*(v/3)) - R/(u*v) = t

That would be the answer I get.