A 200 g, 40-cm-diameter turntable rotates on frictionless bearings at 60 rpm. A 20 g block sits at the center of the turntable. A compressed spring shoots the block radially outward along a frictionless groove in the surface of the turntable. What is the turntable’s rotation angular velocity when the block reaches the outer edge?

mass of turn table = 200 gm = 0.2 kg

Diameter of turntable = 40 cm = 0.4 m

so radius of turn table =0.2 m

mass of block = 20 gm =0.02 kg

rpm = 60.

covert rpm into rad/sec = 2π rad/sec

Use the law of conservation for momentum

initial momentum = final momentum

L_i= L_f..............................................(1)

Get initial momentum

L_i= I_iω_i.............................................(2)

Find moment of inertia

$I_i=\frac{1}{2}(M+m)r^2$

Substitute this in equation(2)

$L_i=\left(\frac{1}{2}(M+m)r^2\right){\omega }_i........................\left(3\right)$

Substitute the given value

$$\begin{gathered} L_i=\left(\frac{1}{2}\times (0.2\mathrm{kg}+0.02\mathrm{kg})(0.2\mathrm{m}{)}^2\right)\times 2\pi {\mathrm{rads}}^{-1} \\ =0.0044{\mathrm{kgm}}^2\times 2\pi {\mathrm{rads}}^{-1}\dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(4\right) \end{gathered}$$

Now find the final momentum

$$\begin{gathered} L_f=I_f{\omega }_f \\ \left(I_t+I_{block}\right){\omega }_f \\ =\left(\frac{1}{2}M{r}^2+m{r}^2\right){\omega }_f\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(5\right) \end{gathered}$$

Substitute the given values

$$\begin{gathered} L_f=\left(\frac{1}{2}\times 0.2\mathrm{kg}\times (0.20\mathrm{m}{)}^2+0.02\mathrm{kg}\times (0.2\mathrm{m}{)}^2\right){\omega }_f \\ =\left(0.0048{\mathrm{kgm}}^2\right){\omega }_f...........................\left(6\right) \end{gathered}$$

Equate (4) and (6) we get

${\omega }_f=\frac{\left(0.0044{\mathrm{kgm}}^2\right)\times 2\pi {\mathrm{rads}}^{-1}}{\left(0.0048{\mathrm{kgm}}^2\right)}=5.76{\mathrm{rads}}^{-1}$