Question

A wheel whose radius is $r$ and moment of inertia about its-own axis is $I$, can rotate freely about its own horizontal axis. A rope is wrapped on the wheel. A boy of mass $m$ is suspended from the free end of the rope. The body is released from rest. The velocity of the body after falling a distance $h$ would be-

• A. ${\left(\frac{mgh}{I}\right)}^{1/2}$
• B. $\left({\frac{2mgh}{m+I}}^{1/2}\right)$
• C. ${\left(\frac{2mgh}{m+I/r^2}\right)}^{1/2}$
• D. ${\left(\frac{m+I}{mgh}\right)}^{1/2}$

Answer 1

The correct option is C ${\left(\frac{2mgh}{m+I/r^2}\right)}^{1/2}$

$\begin{array}{c}\text{* Constraint equation for the boy, of mass}m\\ \Rightarrow mg-T=ma-1\\ \text{* constraint equation for the wheel-}\\ \text{torque}\tau =I_x\text{.}\\ \tau =TRT\to \text{Tension in the string}\end{array}$

$\begin{array}{c}\Rightarrow TR=I\frac{a}{R}\text{sinu}(a=R\alpha )\text{for Ratational motion}\\ T=\frac{I}{R^2}\cdot a-\left(2\right)\\ \text{From eq}D\text{and}\left(2\right)\text{we get}-\\ \text{acceleration}a=\frac{mg{R}^2}{m{R}^2+I}\\ \text{From equation of motion -}\\ \begin{array}{cc}{v}_f^2& =u_0^2+2ah\\ \Rightarrow v_f& ={\left[\frac{2mg{R}^{2}h}{m{R}^2+I}\right]}^{1/2}={\left[\frac{2mgh}{m+I/R^2}\right]}^{1/2}\end{array}\end{array}$