# Homework Help: Time taken to start pure rolling

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1. Oct 2, 2015

### Titan97

1. The problem statement, all variables and given/known data
A solid sphere of radius R is set into motion on a rough horizontal surface with a linear speed v0 in forward direction and angular speed ω0$=\frac{v_0}{2R}$ in counter clockwise direction. Find time after which pure rolling starts.

2. Relevant equations
For pure rolling, $v=\omega R$
$\tau=I\alpha$
$I=\frac{2}{5}mR^2$
$f=\mu N$
(N is normal reaction)

3. The attempt at a solution
Friction acts in backward direction (opposite to velocity) to change the direction of rotation.
First, the angular velocity and linear velocity decreases.
Then the angular velocity becomes zero but the sphere still has a velocity.
Then the angular velocity increases in opposite directions till $\omega=v/R$

$f=\mu mg$
$\alpha=\frac{5\mu g}{2R}$
$a=\mu g$

Time taken for angular velocity to become zero $t_0=\frac{v_0}{5\mu g}$
Velocity at $t_0$ is $v=\frac{4v_0}{5}$
After that, $\omega$ increases from 0 to $v/R$
$\frac{4v_0}{5R}-\frac{\mu gt}{R}=\frac{5\mu g}{2R}t$
This gives $t=\frac{8v_0}{35\mu g}$ which is incorrect. What is the mistake in my approach?

2. Oct 2, 2015

### TSny

Your work looks good. Did you combine the two times to get the total time?

3. Oct 2, 2015

### andrewkirk

You don't state what $t_0$ is but it appears you are intending it to mark the time when the sphere's angular velocity is zero.

That's likely to cause you problems because the question wants the time since the beginning of the experiment, not from the time of zero angular velocity. I think your final answer measures $t$ as time since $t_0$, which is not what the question is asking for. I suggest you re-label, calling the beginning of the experiment $t_0$ (which matches the label $v_0$), the time of zero rotation $t_1$ and the time of commencing rolling $t_2$. The question asks for $t_2-t_0$.

4. Oct 2, 2015

### Titan97

I got the correct answer now @TSny
@andrewkirk , the equation came from $\frac{v}{R}=\omega$.
$v=v'-\mu gt$, $\omega=\alpha t=\frac{5\mu g}{2R}t$
$t$ is measured since $t_0$. I forgot about that.
I could have avoided finding $t_0$ and used two equations $v=v_0-\mu gt$, $\omega=-\omega_0+\alpha t$.

5. Oct 2, 2015

### TSny

OK, good.
You don't need to break the problem up into two parts. The linear and angular accelerations are constant during the whole time. So, you can set up the equations to take you all the way from initial conditions to the final time.

6. Oct 3, 2015

### andrewkirk

It's interesting to note that the question assumes, with no justification, that the reader will interpret 'forward direction' to mean 'from left to right'. That assumption turned out to be accurate in this case, with the three people that looked at the problem all assuming it meant from left to right.

I wonder though, whether in a country where writing goes from right to left - such as in Arabic, Urdu or Farsi - it would typically be interpreted as meaning the sphere was moving from right to left, in which case the answer would be a much shorter time.

7. Oct 3, 2015

### Titan97

A picture is given. But its small.