This problem aims to familiarize us with objects **moving** in a **circular path.** The concepts required to solve this problem include **angular velocity, right-hand rule**, and **angular momentum.**

In physics, **angular velocity** is the measure of the **rotation** of an object in a specific time period. In simple words, it is the **rate** at which an **object spins** around an axis. It is denoted by the Greek letter $\omega$ and its **formula** is:

\[ \omega = \dfrac{\phi}{t}\]

Where $\phi$ is the **angular displacement** and $t$ is the change in **time** to cover that distance.

A**ngular momentum** is the property of a **revolving** object which is given by the moment of **inertia** into the **angular** velocity. The **formula** is:

\[ \vec{L} = I\times \vec{\omega} \]

Where $I$ is the **rotational inertia,** and $\vec{\omega}$ is the **angular velocity.**

## Expert Answer

As per the **statement,** we are given the following **information:**

The **mass** of the turntable $M = 2 kg$,

**Diameter** of the turntable $d = 20cm =0.2m$,

**Initial Angular velocity** $\omega = \dfrac{100rev}{minute} = 100\times \dfrac{2\pi}{60} = 10.47\space rad/s$,

And the **mass** of the **two** blocks $m = 500g = 0.5 kg$.

To find the **angular velocity** of the turntable, we will **apply** the principle of **conservation** of **momentum,** since they change the moment of **inertia** of the whole system when they **stick** with each other. Thus, the **angular velocity** of the system changes.

By using the **the ****conservation** of momentum principle:

\[L_{initial}=L_{final}\]

\[ I_{turntable}\times\omega = I_{block_1} \omega^{‘}+I_{turntable}\omega^{‘} + I_{block_2}\omega^{‘} \]

Where $\omega^{‘}\neq\omega $ i.e. the **angular velocity.**

Solving for $\omega^{‘} $, gives us:

\[\omega^{‘}=\dfrac{I_{turntable} \omega}{I_{block_1}+I_{turntable} + I_{block_2}}\]

Let’s first find the **two possible** unknowns:

\[ I_{turntable}=M\dfrac{r^2}{2}\]

\[ I_{turntable}=2\dfrac{0.1^2}{2} = 0.01\]

\[ I_{block_1}=mr^2 0.5 \times 0.1^2\]

\[ I_{block_1}=0.005 = I_{block_2} \]

**Plugging** the values gives us:

\[\omega^{‘}=\dfrac{0.01\times 10.47}{0.005 + 0.01 + 0.005} \]

\[\omega^{‘} = 5.235\space rad/s \]

\[\omega^{‘} = 5.235\times \dfrac{60}{2\pi} rev/min \]

\[\omega^{‘} = 50\space rev/min\]

## Numerical Result

The turntable’s **angular velocity** in rpm is calculated as $\omega^{‘} = 50\space rev/min$.

## Example

A $10 g$ **bullet** with speeds of $400 m/s$ hits a $10 kg$, $1.0 m$ broad **door** at the corner opposite the hinge. The **bullet** entrenches itself in the **door,** forcing the door to swing open. Find the **angular velocity** of the door just after the hit?

The **initial angular momentum** is retained completely inside the bullet. So the **angular momentum** before the impact will be:

\[ (M_{bullet})×(V_{bullet})×(distance)\]

\[ = (M_{bullet})(V_{bullet})(R)\]

Where $R$ is the width of the door.

The **final angular momentum** includes rotating objects, so it’s suitable to represent it as angular speed $\omega$.

So the **angular momentum** after the bullet hits is:

\[ \omega\times I\]

\[=\omega (I_{door} + I_{bullet})\]

**Moment** of **inertia** for the **door** is $I = \dfrac{1}{3}MR^2$,

The **moment** of **inertia** for the **bullet** is $I = MR^2$.

The **equation** becomes:

\[ \omega(\dfrac{1}{3}(M_{door})R^2 + (M_{bullet})R^2)\]

Using principle of **angular momentum:**

\[(M_{bullet})(V_{bullet})(R) = \omega(\dfrac{1}{3}(M_{door})R^2 + (M_{bullet})R^2)\]

Thus:

\[\omega = \dfrac{(M_{bullet})(V_{bullet})(R)}{\dfrac{1}{3}(M_{door})R^2 + (M_{bullet})R^2)}\]

\[= \dfrac{(M_{bullet})(V_{bullet})}{(R(\dfrac{M_{door}}{3} + M_{bullet})})\]

\[= \dfrac{(10g)(400m/s)}{(1.0m(\dfrac{10kg}{3} + 10kg)})\]

\[= 1.196 rad/sec\]