## Related questions with answers

Particles of mud are thrown from the rim of a rolling wheel. If the forward speed of the wheel is $v_{0}$, and the radius of the wheel is b, show that the greatest height above the ground that the mud can go is $b+\frac{v_{0}^{2}}{2 g}+\frac{g b^{2}}{2 v_{0}^{2}}$ At what point on the rolling wheel does this mud leave? (Note: It is necessary to assume that $v_{0}^{2} \geq b g$.)

Solution

VerifiedLet us take a point on the lower part of the rim relative to center to be

$p = b\cos\theta\textbf{i} - b\sin\theta\textbf{j}$

We can find velocity from this p

$v = -v_o\sin\theta\textbf{i} - v_o\cos\theta\textbf{j}$

As this is relative to the center velocity, the velocity related to the ground (as $v_o$ is in x direction as given) will be

$v = v_o\textbf{i} - v_o\sin\theta\textbf{i} - v_o\cos\theta\textbf{j}$

$v = v_o(1 - \sin\theta)\textbf{i} - v_o\cos\theta\textbf{j}$

Let us now calculate the height using this velocity in vertical direction.

We know that we the mud will be already at a height from ground which is $-b\sin\theta$, thus the total height will be

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