A 2.0 kg, 20cm-diameter turntable rotates at 100 rpm on friction-less bearings. Two 500 g 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?

What Is The TurntableS Angular Velocity In Rpm Just After This Event

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.

Circular path

Circular path

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.

Angular 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.

Angular momentum

Angular momentum

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:


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

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 $\omega^{‘} = 50\space rev/min$.


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:

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


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

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

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

\[= 1.196 rad/sec\]

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