Jumping Into/Out of Moving Disc: Videos & Practice Problems

In angular collision problems, the conservation of angular momentum is key. For a disc with mass \(M = 200 \, \text{kg}\) and radius \(R = 4 \, \text{m}\), the moment of inertia \(I\) is calculated as \(I = \frac{1}{2}MR^2 = 1600 \, \text{kg} \cdot \text{m}^2\). When a person with mass \(m = 80 \, \text{kg}\) jumps onto the disc, the new angular speed \(\omega_f\) can be determined using \(L_{\text{initial}} = L_{\text{final}}\). Different scenarios, such as jumping with negligible speed or tangentially, affect the final angular speed, demonstrating the principles of angular momentum and rotational dynamics.

Jumping into a moving disc

In angular collision problems, understanding the conservation of angular momentum is crucial. Consider a disc with a mass of 200 kg and a radius of 4 meters. The moment of inertia (I) for a disc is calculated using the formula:

I = \frac{1}{2} m r^2

Substituting the values, we find:

I = \frac{1}{2} \times 200 \, \text{kg} \times (4 \, \text{m})^2 = 1600 \, \text{kg m}^2

The disc initially spins with an angular velocity (ω) of 2 radians per second. When an 80 kg person jumps onto the disc with negligible speed, we can analyze the change in angular velocity using the conservation of angular momentum:

L_{initial} = L_{final}

Initially, the angular momentum of the disc is:

L_{initial} = I_{disc} \cdot \omega_{initial} = 1600 \, \text{kg m}^2 \cdot 2 \, \text{rad/s} = 3200 \, \text{kg m}^2/s

When the person enters the disc, their contribution to angular momentum is zero since they have negligible speed. Thus, the final angular momentum is:

L_{final} = (I_{disc} + I_{person}) \cdot \omega_{final}

To find the moment of inertia of the person treated as a point mass at the edge of the disc:

I_{person} = m_{person} \cdot r^2 = 80 \, \text{kg} \cdot (4 \, \text{m})^2 = 1280 \, \text{kg m}^2

Now, substituting into the conservation equation:

3200 = (1600 + 1280) \cdot \omega_{final}

Solving for ω_final gives:

\omega_{final} = \frac{3200}{2880} \approx 1.11 \, \text{rad/s}

This indicates that the disc slows down due to the added mass. In the next scenario, if the person jumps into the disc at its edge with a speed of 9 m/s directed towards the center, they do not contribute any angular momentum, similar to the first case. Thus, the final angular velocity remains:

\omega_{final} = 1.11 \, \text{rad/s}

In contrast, if the person jumps tangentially with a speed of 9 m/s, the situation changes. Here, the angular momentum contribution from the person must be considered. The initial angular momentum remains the same, but the person’s tangential velocity will affect the final angular velocity. For part c, where the jump is against the disc's rotation, the velocity is negative:

L_{initial} = 3200

L_{final} = (1600 + 1280) \cdot \omega_{final} - 80 \cdot 9 \cdot 4

Solving this gives:

\omega_{final} \approx 0.11 \, \text{rad/s}

For part d, where the jump is in the same direction as the rotation, the velocity is positive:

L_{final} = (1600 + 1280) \cdot \omega_{final} + 80 \cdot 9 \cdot 4

Solving this results in:

\omega_{final} \approx 2.11 \, \text{rad/s}

These calculations illustrate how the direction and speed of the person jumping onto the disc significantly affect the final angular velocity. Adding mass always reduces the angular velocity, but the direction of the jump can either counteract or enhance the disc's rotation.

A 200 kg disc 2 m in radius spins around a perpendicular axis through its center, with a person on it, at 3 rad/s counter-clockwise. The person has mass 70 kg, is at rest (relative to the disc, that is, spins with it) at the disc's edge, and can be treated as a point mass. If the person jumps tangentially out of the disc with 10 m/s (relative to the floor), as shown by the red arrow, what new angular speed will the disc have as a result? If the person steps out into ice with negligible speed of his/her own, what speed would it have upon exiting?

• A. 1.6 rad/s
• B. 3.2 rad/s
• C. 5.8 rad/s
• D. 8.6 rad/s

Answer: (A)

Here’s what students ask on this topic:

What is the moment of inertia for a disc and how is it calculated?

The moment of inertia for a disc is calculated using the formula $I=\frac{1}{2}m{r}^2$, where $m$ is the mass of the disc and $r$ is its radius. This formula is applicable to solid discs and cylinders, reflecting how mass is distributed relative to the axis of rotation. The moment of inertia is a measure of an object's resistance to changes in its rotational motion. For a disc spinning around a perpendicular axis through its center, this formula helps determine how the disc will respond to external forces or masses, such as a person jumping onto it.

How does jumping onto a moving disc affect its angular speed?

Jumping onto a moving disc affects its angular speed due to the conservation of angular momentum. When a person jumps onto the disc, they add mass and potentially momentum, altering the disc's rotational motion. The final angular speed can be calculated using the conservation equation: $I_1{\omega }_1+I_2{\omega }_2=I_1{\omega }_1+I_2{\omega }_2$. The direction of the jump influences whether the disc speeds up or slows down. Jumping tangentially in the direction of rotation increases speed, while jumping against it decreases speed.

What is the effect of jumping tangentially onto a disc?

Jumping tangentially onto a disc affects its angular speed by contributing to its rotational motion. If the jump is in the direction of the disc's rotation, it adds positive angular momentum, increasing the disc's speed. Conversely, if the jump is against the rotation, it adds negative angular momentum, slowing the disc down. The effect is calculated using the conservation of angular momentum, where the initial and final angular momenta are equated. The mass and velocity of the jumper, along with the radius of the disc, determine the magnitude of the change in angular speed. This principle illustrates how external forces and mass distribution influence rotational dynamics.

How does the conservation of angular momentum apply to a person jumping onto a disc?

The conservation of angular momentum applies to a person jumping onto a disc by ensuring that the total angular momentum before and after the jump remains constant. Initially, the disc has its own angular momentum, calculated as $I{\omega }_{\mathrm{initial}}$. When the person jumps onto the disc, their mass and velocity contribute additional angular momentum, calculated as $mvr$. The final angular speed is determined by equating the initial and final angular momenta, considering both the disc and the person. This principle highlights how mass and velocity influence rotational motion, maintaining the system's angular momentum.

What happens when a person jumps onto a disc with negligible speed?

When a person jumps onto a disc with negligible speed, they do not contribute any linear momentum to the disc's rotation. The only effect is the addition of mass, which increases the moment of inertia and consequently slows down the disc's angular speed. The final angular speed is calculated using the conservation of angular momentum, where the initial angular momentum of the disc is equated to the final angular momentum of the combined system. Since the person does not add any velocity, the change in angular speed is solely due to the increased mass, demonstrating how mass distribution affects rotational dynamics.