Question

A wheel of radius $r$ and moment of inertia $I$ about its axis is fixed at the top of an inclined plane of inclination $\theta$ as shown in the figure. A string is wrapped round the wheel and its free end supports a block of mass $M$ which can slide on the plane. Initially, the wheel is rotating at a speed of $\omega$ in a direction such that the block slides up the plane. How far will the block move before stopping?

• A. $\frac{(I+M{r}^3){\omega }^2}{4Mg\sin \theta }$
• B. $\frac{(I+M{r}^2){\omega }^2}{2Mg\sin \theta }$
• C. $\frac{(I+M{r}^2){\omega }^2}{5Mg\sin \theta }$
• D. $\frac{(I+M{r}^2){\omega }^2}{7Mg\sin \theta }$

Answer 1

The correct option is B $\frac{(I+M{r}^2){\omega }^2}{2Mg\sin \theta }$

$△u=mglsin\theta$

MOI of system $=(I+m{r}^2)$

Also initial kinetic energy =$△u$

$\therefore \frac{1}{2}(I+m{r}^2){\omega }^2=mglsin\theta \Rightarrow l=\frac{(I+m{r}^2){\omega }^2}{2mgsin\theta }$