A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear spe

Question

A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ( $P$ and $Q$ are any highest and lowest points on the wheel respectively)?

Professor PC Thomas Classes · Answer

(b) : For pure rolling $v_{c m}=r \omega$ Given, $v_{c m}=v$ \ $$ \begin{equation*} 	herefore \quad v=r \omega 	ag{i} \end{equation*} $$ Net velocity is the vector sum of translational and rotational velocity and $	heta$ is the angle between translational and rotational velocity. $$ \left|\vec{v}_{n e t}ight|=\sqrt{v^2+(r \omega)^2+2(v)(r \omega) \cos 	heta} $$ For point $Q, 	heta=180^{\circ}$ $$ \begin{aligned} \left|\vec{v}_Qight| & =\sqrt{v^2+r^2 \omega^2+2 r v \omega \cos 180^{\circ}} \ & =\sqrt{(v-r \omega)^2}=v-r \omega \end{aligned} \ \ \left|\vec{v}_Qight|=r \omega-r \omega=0 \quad(	ext { using }(\mathrm{i})) 	ag{ii} \ $$ Now for point $P, 	heta=0^{\circ}$ Using eq., (i) we have $ \begin{aligned} & \left|\vec{v}_Pight|=\sqrt{v^2+(r \omega)^2+2(v)(r \omega) \cos 0^{\circ}} \ & =\sqrt{(v+r \omega)^2}=v+r \omega \end{aligned} $ \ Using eq. (i), we have $ \left|\vec{v}_Pight|=2 r \omega $ \ So, it is clear from equation (ii) and (iii) that point $P$ moves faster than point $Q$.