Question

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

• A. $u^{2\:}/\:2g\:$
• B. $\frac{u^2}{2g}+\frac{gR^2}{2u^2}$
• C. $0$
• D. $R+\frac{u^2}{2g}$

Answer 1

Answer: (B)

Let a mud blob is detached from the circumference of wheel with initial speed $u$ at angle $\theta$ as shown below.

Height upto which mud blob can be thrown is
$h =$ Maximum height of projectile $+$ Height at which mud blob thrown is
$\therefore h=\frac{u^{2}\sin^{2}\theta}{2g}+R \cos \theta \ldots\left(i\right)$
$h$ is maximum when $\frac{dh}{d\theta}=0$
$\Rightarrow \frac{d}{d\theta} \left(\frac{u^{2} \sin^{2}\theta}{2g}+R \cos \theta\right)$
$\Rightarrow \frac{u^{2}}{2g}\cdot 2\sin \theta \cos \theta-R \sin \theta=0$
$\Rightarrow \frac{u^{2}}{g}\cos\theta =R$ or $\cos\theta=\frac{Rg}{u^{2}}$
$\Rightarrow \sin^{2}\theta=1-\frac{R^{2}g^{2}}{u^{2}}$
Substituting these values in Eq $\left(i\right),$ we get
$\lambda_{\max} =\frac{u^{2}}{2g}\left(1-\frac{R^{2}g^{2}}{u^{4}}\right)+R\left(\frac{Rg}{u^{2}}\right)$
$=\frac{u^{2}}{2g}-\frac{R^{2}g}{2u^{2}}+\frac{R^{2}g}{u^{2}}=\frac{u^{2}}{2g}+\frac{R^{2}g}{2u^{2}}$