# Homework Help: Mass and ring

1. Sep 16, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
This problem involves solving a simple differential equation. A block of mass $m$ slides on a frictionless table. It is constrained to move inside a ring of radius $l$ that is fixed to the table. At t = 0, the block is moving along the inside of the ring (in the tangential direction) with velocity $v_0$. The coefficient of friction between the block and the ring is $\mu$. Find the velocity of the block at later times

2. Relevant equations

3. The attempt at a solution
So we start by simply identifying the forces on the body in the plane of rotation (since the normal force on the table cancels with gravitational force). So we have the normal force on the block from the ring, and we have the frictional force between the block and the ring. Using polar coordinates, we find that

$-N_r = m(\ddot{r} - r \dot{\theta}^2)$
$\ddot{r} = 0$
Thus
$N_r = mr \dot{\theta}^2$

In the tangential direction:
$-f_{friction} = m(r \ddot{\theta} + 2 \dot{r} \dot{\theta})$
$\dot{r} = 0$
Thus
$f = -mr \ddot{\theta}$

Since we have kinetic friction, $f = \mu_k N_r$

So

$\mu m r \dot{\theta}^2 = mr \ddot{\theta}$

$\ddot{\theta} + \mu \dot{\theta}^2 = 0$

However, this is a nonlinear differential equation, so it can't be the answer. What am I doing wrong?

2. Sep 16, 2016

### kuruman

Suppose you let ω = dθ/dt. Substitute and get a diff. eq. in ω. Can you solve that one first?

3. Sep 16, 2016

### Mr Davis 97

So if I do as you say, would I get $\displaystyle \frac{1}{\omega} - \frac{1}{\omega_0} = -\mu t$? To get velocity explicitly would I make the substitution $\omega = \frac{v}{r}$?

4. Sep 16, 2016

### kuruman

Yep.

5. Sep 16, 2016

### Mr Davis 97

Cool, thanks! I made a big deal out of nothing

6. Sep 16, 2016

### Mr Davis 97

One more thing, if I find $\theta$, which I found to be $\displaystyle \frac{1}{\mu} \ln|\mu t + \frac{1}{\omega_0}|$, how would I then find the position for any time t?

7. Sep 16, 2016

### kuruman

What position are you talking about? Here, the mass is going around in a circle therefore the only position that makes sense is the angular position as a function of time. That's θ(t) which you have attempted to find. I say "attempted" because your expression is incorrect. The argument of the logarithm must be dimensionless and yours is not.

Last edited: Sep 16, 2016
8. Sep 16, 2016

### Mr Davis 97

$\displaystyle -\frac{1}{\omega} + \frac{1}{\omega_0} = -\mu t$
$\displaystyle \frac{1}{\omega} = \mu t + \frac{1}{\omega_0}$
$\displaystyle \omega = \frac{1}{\mu t + \frac{1}{\omega_0}}$
$\displaystyle \frac{d \theta}{dt} = \frac{1}{\mu t + \frac{1}{\omega_0}}$
$\displaystyle \int_{\theta_0}^{\theta} d \theta = \int_0^t \frac{1}{\mu t + \frac{1}{\omega_0}} dt$
$\displaystyle \theta - \theta_0 = \frac{1}{\mu} \ln|\mu t + \frac{1}{\omega_0}|$

What am I doing wrong?

9. Sep 16, 2016

### kuruman

Forget the absolute value of the argument of the log. Just evaluate the log at the upper limit and subtract from it the log evaluated at the lower limit. Then observe that the difference of the two logs is the log of the ratio of the arguments, a dimensionless quantity.

10. Sep 16, 2016

### Mr Davis 97

Can't believe I didn't see that. So the answer would be $\theta = ln(1 + \omega_0 \mu t)$?

11. Sep 16, 2016

### kuruman

It would be that.