I have this problem here:
I think I know how to solve everything except for the number 2). I studied in the mechanics subject that $W_{ext}=K_{2}-K_{1}$. Where $W_{ext}$ is the external forces' work and $K_2$ and $K_1$ are the Kinetic energies in the moment 2 and 1, respectively.
For 3) I suppose I can use :
$W_{ext}=-mgh$
and $K_1=\frac{1}{2}mV^2+\frac{1}{2}I\dot{\theta}'^2=\frac{1}{2}m(r\dot{\theta}')^2+\frac{1}{2}(\frac{1}{2}mr^2)\dot{\theta}'^2=\frac{3}{4}mr^2\dot{\theta}'^2$
Equally, $K_2=\frac{3}{4}mr^2\dot{\theta}''^2$
Therefore $-mgh=\frac{3}{4}mr^2\dot{\theta}''^2-\frac{3}{4}mr^2\dot{\theta}'^2\rightarrow \dot{\theta}''=\sqrt{\frac{\frac{3}{4}mr^2\dot{\theta}'^2-mgh}{\frac{3}{4}mr^2}}=\sqrt{\dot{\theta}'^2-\frac{4gh}{3r^2}}$
And for 4), I can just equal the kinetic energy to the potential energy necessary for climbing the step, so $\frac{3}{4}mr^2\dot{\theta}_{min}'^2=mgh\rightarrow \dot{\theta}_{min}'=\sqrt{\frac{4gh}{3r^2}}$. Which makes sense comparing it to the solution in 3).
Are the approaches I used in this two parts correct? Because when I try to use this in 2), as just after it collided there is not a change in the height and therefore the $W_{ext}=0$, and that just at the colission the velocity is $0$, $\dot{\theta}'=3\dot{\theta}$, which doesn't make sense for me. Could you help me with this?
Thank you very much
