Question

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm onfrictionless bearings. Two 500 g blocks fall from above, hitthe turntable simultaneously atopposite ends of a diagonal, andstickWhat is the turntable's angular velocity, inrpm, just after this event?

Answer 1

Concepts and reason

The concepts required to solve the given question is moment of inertia of the disk and conservation of angular momentum.

Initially, calculate the moment of inertia of the disk (turntable). Later, find the moment of inertia of the disk after two blocks hits and sticks to the edges of the turn table. Finally, apply the law of conservation of angular momentum to find the final angular velocity of the system of turntable with two blocks at its edge.

Fundamentals

The expression moment of inertia of the turntable is as follows:

${I_1} = \frac{1}{2}M{R^2}$

Here, M is the mass of the turntable and R is the radius of the turntable.

The expression for the moment of inertia of the each block that felt and attached at the ends of the turntable is,

${I_0} = m{R^2}$

Here, $m$is the mass of the block.

The conservation of angular momentum states that the total angular momentum of the system that is in rotational equilibrium is constant. The expression for the conservation of the angular momentum is as follows:

${I_1}{\omega _1} = {I_2}{\omega _2}$

Here, ${I_1}$and ${I_2}$ are the initial and final moment of inertias of the turn table, ${\omega _1}$ and ${\omega _2}$ are the initial and final angular velocities of the turntable.

The relation between the radius and the diameter is as follows:

$R = \frac{D}{2}$

Here, D is the diameter.

Substitute 20 cm for D.

$\begin{array}{c}\\R = \frac{{20{\rm{ cm}}}}{2}\\\\ = \frac{{20{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.00{\rm{ cm}}}}} \right)}}{2}\\\\ = 0.1{\rm{ m}}\\\end{array}$

Substitute 2.0 kg for M and $0.1{\rm{ m}}$for R in the equation${I_1} = \frac{1}{2}M{R^2}$.

$\begin{array}{c}\\{I_1} = \frac{1}{2}\left( {2.0{\rm{ kg}}} \right){\left( {0.1{\rm{ m}}} \right)^2}\\\\ = 0.01{\rm{ kg}} \cdot {{\rm{m}}^2}\\\end{array}$

The moment of inertia of each block is,

${I_0} = m{R^2}$

Total moment of inertia of two blocks is,

${I'_0} = 2m{R^2}$

The final moment of inertia of the turn table is,

${I_2} = {I_1} + {I'_0}$

Replace ${I'_0}$with$2m{R^2}$.

${I_2} = {I_1} + 2m{R^2}$

Substitute $0.01{\rm{ kg}} \cdot {{\rm{m}}^2}$ for${I_1}$, 500 g for M, and 0.1 m for R in the equation${I_2} = {I_1} + 2M{R^2}$.

$\begin{array}{c}\\{I_2} = 0.01{\rm{ kg}} \cdot {{\rm{m}}^2} + 2\left( {500{\rm{ g}}\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1.00{\rm{ g}}}}} \right)} \right){\left( {0.1{\rm{ m}}} \right)^2}\\\\ = 0.02{\rm{ kg}} \cdot {{\rm{m}}^2}\\\end{array}$

Rearrange the equation ${I_1}{\omega _1} = {I_2}{\omega _2}$ for${\omega _2}$.

${\omega _2} = \frac{{{I_1}{\omega _1}}}{{{I_2}}}$

Substitute$0.01{\rm{ kg}} \cdot {{\rm{m}}^2}$${I_1}$, 100 rpm for${\omega _1}$, and $0.02{\rm{ kg}} \cdot {{\rm{m}}^2}$ for ${I_2}$ in the equation ${\omega _2} = \frac{{{I_1}{\omega _1}}}{{{I_2}}}$.

$\begin{array}{c}\\{\omega _2} = \frac{{\left( {0.01{\rm{ kg}} \cdot {{\rm{m}}^2}} \right)\left( {100{\rm{ rpm}}} \right)}}{{0.02{\rm{ kg}} \cdot {{\rm{m}}^2}}}\\\\ = 50{\rm{ rpm}}\\\end{array}$

Ans:

The magnitude of the turntable’s angular velocity is 50 rpm.