Question

A V $2.0\mathrm{kg},20-\mathrm{cm}$-diameter turntable rotates at $100rpm$on frictionless bearings. Two $500g$blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable's angular velocity, in rpm, just after this event?

Short answer

Therefore, the final angular velocity of the turntable is$50rpm$.

Step-by-step solution

Step :1 Introduction 

According to conservation of angular momentum, the initial angular momentum of the object is equal to final angular momentum of the object.

$L_i=L_f$

Here, $L_1$is the initial angular momentum of the object and $L_f$is the final angular momentum of the object.

Step :2 Explanation

The initial angular momentum of the turntable

$L_i=I_i{\omega }_i$

Here $L_j$is the initial moment of inertia, ${\omega }_i$is initial angular velocity. The final angular momentum of the turntable is,

$L_f=I_f{\omega }_f$

Here, is the final moment of inertia and ${\omega }_y$is final angular velocity.

Substitute $I_f{\omega }_f\text{for}{L}_{\overline{c}}\text{and}{I}_f{\omega }_f\text{for}{L}_f$in the equation $L_{\mathrm{i}}=L_f$and solve for ${\omega }_f$.

$\begin{array}{r}{I}_f{\omega }_i=I_f{\omega }_f\\ {\omega }_f=\left(\frac{I_f}{I_f}\right){\omega }_i\end{array}$

The initial momentum of the inertia in turntable is

$I_i=\frac{1}{2}M{r}^2$

Here $M$is mass of turntable and$R$is radius of turntable.

The final angular momentum is due to turntable and two blocks. The final moment of inertia of the system is,

$I_f=\frac{1}{2}M{r}^2+2m{r}^2$

Here, $m$ is the mass of each block and $r$ is the distance of block from the center of the turntable.

Step  : 3 Relationship of radius and diameter 

The relation between radius and diameter on the turntable is

$r=\frac{d}{2}$

Here $d$is the diameter of the turntable

Substitute $20cm$for $d$

$r=\frac{20\mathrm{cm}}{2}$

$=10\mathrm{cm}\left(\frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)$

$0.1m$

$\begin{array}{r}r=\frac{20\mathrm{cm}}{2}\\ =10\mathrm{cm}\left(\frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)\\ =0.1\mathrm{m}\end{array}$

Step :4 Substitution

Substitute $\frac{1}{2}M{r}^2\text{for}{I}_i\text{and}\frac{1}{2}M{r}^2+2m{r}^2\text{for}{I}_r$in equation ${\omega }_f=\left(\frac{I_i}{I_f}\right){\omega }_i$and solve ${\omega }_f$

${\omega }_f=\left(\frac{\frac{1}{2}M{r}^2}{\frac{1}{2}M{r}^2+2m{r}^2}\right){\omega }_1$

Substitute $2.0\mathrm{kg}\text{for}M,0.1\mathrm{m}\text{for}r,500\mathrm{g}\text{for}m\text{and}100\mathrm{rpm}\text{for}{\omega }_{\mathrm{p}}\text{.}$

${\omega }_f=\left(\frac{\frac{1}{2}\left(2.0\mathrm{kg}\right)(0.1\mathrm{m}{)}^2}{\frac{1}{2}\left(2.0\mathrm{kg}\right)(0.1\mathrm{m}{)}^2+2(500\mathrm{g}\left)\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{g}}\right)\right(0.1\mathrm{m}{)}^2}\right)\left(100\mathrm{rpm}\right)$

=50rpm