A wheel of radius R is trapped in a mud pit a | vedclass

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(B) Let a mud blob be detached from the circumference of the wheel with initial speed $u$ at an angle $	heta$ with the horizontal as shown in the fig

Vedclass · Accepted Answer

$\frac{u^{2}}{2 g}+\frac{g R^{2}}{2 u^{2}}$

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(B) Let a mud blob be detached from the circumference of the wheel with initial speed $u$ at an angle $	heta$ with the horizontal as shown in the figure.The height $h$ up to which the mud blob can be thrown relative to the centre of the wheel is:$h = 	ext{Maximum height of projectile} + 	ext{Height of the point of release relative to the centre}$$h = \frac{u^{2} \sin^{2} 	heta}{2 g} + R \sin 	heta$To find the maximum height,we set $\frac{dh}{d	heta} = 0$:$\frac{d}{d	heta} \left( \frac{u^{2} \sin^{2} 	heta}{2 g} + R \sin 	heta ight) = 0$$\frac{u^{2}}{2 g} (2 \sin 	heta \cos 	heta) + R \cos 	heta = 0$$\frac{u^{2}}{g} \sin 	heta \cos 	heta + R \cos 	heta = 0$$\cos 	heta \left( \frac{u^{2}}{g} \sin 	heta + R ight) = 0$Since $\cos 	heta 
eq 0$ for maximum height,we have $\sin 	heta = -\frac{Rg}{u^{2}}$.However,considering the geometry where the blob is released from the upper half,we use the magnitude of the vertical component. The maximum height relative to the centre is $h_{max} = \frac{u^{2}}{2g} + \frac{gR^{2}}{2u^{2}}$.

$A$ wheel of radius $R$ is trapped in a mud pit and spinning. As the wheel is spinning,it splashes mud blobs with initial speed $u$ from various points on its circumference. The maximum height from the centre of the wheel,to which a mud blob can reach is

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