Ball rolling up a ramp with friction -- Find max height (Unsolved)

[srmico]

I ended up solving it this way, if anyone is interested:
From the conservation of energy we have that
\begin{equation}
\frac{1}{2}mv^{2}=mgh+\Delta E
\label{eq:1}
\end{equation}
where $\Delta E$ is the energy difference due to friction force.
\begin{equation*}
\Delta E=E_{f}\text{(energy lost due to friction)}-E_{r}\text{(rotational energy)}
\end{equation*}
To calculate $E_{f}$ we have
\begin{equation}
F_{f}=ma=-\mu mg\cos(\theta)
\label{eq:2}
\end{equation}
\begin{equation}
E_{f}=-\mu mg\cos(\theta)d
\label{eq:3}
\end{equation}
Now from \ref{eq:2} we find
\begin{equation*}
a=-\mu g\cos(\theta)
\end{equation*}
And integrating with respect to time we find the velocity
\begin{equation*}
v=-\mu g\cos(\theta)t+v_{0}
\end{equation*}
To be able to find d, we need the time where the rolling condition is fulfilled i.e.
\begin{equation}
w(t_0)=\frac{v(t_0)}{r}
\label{eq:4}
\end{equation}
We are missing an expression for $\omega$, but we can get one from
\begin{equation*}
\tau = I\alpha = F_{f}r= \mu mg\cos(\theta)r
\end{equation*}
Solving fro $\alpha$ and using $I=\frac{2}{5}mr^{2}$
\begin{equation*}
\alpha = \frac{5}{2}\frac{\mu g \cos(\theta)}{r}
\end{equation*}
Integrating with respect to time we find
\begin{equation*}
\omega = \frac{5}{2}\frac{\mu g \cos(\theta)}{r}t
\end{equation*}
Now that we have an equation for $\omega$ and an equation for $v$ we can substitute in \ref{eq:4} to find time $t_{0}$ at which the rolling condition is met. Solving for $t_{0}$ we find
\begin{equation*}
t_{0} = \frac{2V_{0}}{7\mu g \cos(\theta)}
\end{equation*}
Since we have an expression for $v(t)$, we can integrate from $t=0$ to $t=t_{0}$ to get the distance traveled in that interval
\begin{equation*}
d=\int_{0}^{\frac{2V_{0}}{7\mu g \cos(\theta)}}(-\mu g\cos(\theta)t+v_{0}) dt= \frac{12V_{0}^{2}}{49\mu g\cos(\theta)}
\end{equation*}
Substituting $d$ at \ref{eq:3} we finally get
\begin{equation*}
E_{f}=\frac{-12}{49}mV_{0}^{2}
\end{equation*}
Now we need an expression for $E_{r}$ to see how much energy was stored as rotational energy. For that we have
\begin{equation*}
E_{r}=\frac{I\omega(t_{0})^{2}}{2}=\frac{5}{49}mV_{0}^{2}
\end{equation*}
Finally, substituting $E_{f}$ and $E_{r}$ in \ref{eq:1}
\begin{equation*}
\frac{1}{2}mv^{2}=mgh+(E_{f}-E_{r})=mgh+\frac{7}{49}mV_{0}^{2}
\end{equation*}
And solving for $h$
\begin{equation*}
h=\frac{5V_{0}^{2}}{14g}
\end{equation*}

 

[Doc Al]

A few comments.

From the conservation of energy we have that
$$\frac{1}{2}mv^{2}=mgh+\Delta E$$
where $\Delta E$ is the energy difference due to friction force.
$$\Delta E=E_{f}\text{(energy lost due to friction)}-E_{r}\text{(rotational energy)}$$

To apply conservation of energy you'd need something like: KE_i = PE_f + KE_f + rotKE_f + Energy lost.

To calculate $E_{f}$ we have

$$F_{f}=ma=-\mu mg\cos(\theta)$$

$$E_{f}=-\mu mg\cos(\theta)d$$

You have calculated the friction force times the distance the ball moved, but that is not the work done by friction, as the ball rolls as well as slides. (It's the pseudowork or center of mass work done by friction, but that's something different.)

Now from 2\ref{eq:2} we find

​

$$a=-\mu g\cos(\theta)$$

That would be the acceleration if friction were the only force. What about gravity?

 

[srmico]

I also think it's wrong, but that's what my teacher assistant told me to do. I asked him about the gravity part, and he said that it's inside the "mgh" (I know it's wrong). As I said, this is from last semester, so I will not spend any more time trying to solve it, because I don't have it (I have exams about electricity and magnetism next week).
Thanks a lot for all the answers :D

 

[Doc Al]

I asked him about the gravity part, and he said that it's inside the "mgh" (I know it's wrong).

When doing energy calculations, the work done by gravity is included in the mgh. But not for calculating the acceleration.

Good luck on your exams!