Answer
The mud is at 9 o'clock position. The velocity of this point will be the resultant of velocity towards the center and $r\omega$.
So, $v=\sqrt (v_{cm}^{2}+r\omega^{2})$
For pure rolling, $v_{cm}=r\omega$,
So, $v=\sqrt 2v_{cm}$
Hence, $\theta=tan^{-1}\frac{r\omega}{v_{cm}}=45^{\circ}$
Hence the mud on the wheel will fly off at an angle of $45^{\circ}$ with horizontal and with a speed equal to $\sqrt 2v_{cm}$.
Work Step by Step
The mud is at 9 o'clock position. The velocity of this point will be the resultant of velocity towards the center and $r\omega$.
So, $v=\sqrt (v_{cm}^{2}+r\omega^{2})$
For pure rolling, $v_{cm}=r\omega$,
So, $v=\sqrt 2v_{cm}$
Hence, $\theta=tan^{-1}\frac{r\omega}{v_{cm}}=45^{\circ}$
Hence the mud on the wheel will fly off at an angle of $45^{\circ}$ with horizontal and with a speed equal to $\sqrt 2v_{cm}$.
