Problem 61 - University Physics with Modern Physics

Question

A 25.0 -kg boy stands \(2.00 \mathrm{~m}\) from the center of a frictionless playground merry-go-round, which has a moment of inertia of \(200 . \mathrm{kg} \mathrm{m}^{2} .\) The boy begins to run in a circular path with a speed of \(0.600 \mathrm{~m} / \mathrm{s}\) relative to the ground. a) Calculate the angular velocity of the merry-go-round. b) Calculate the speed of the boy relative to the surface of the merry-go- round.

Step-by-Step Solution

Answer

Answer: The angular velocity of the merry-go-round is 0.15 rad/s, and the speed of the boy relative to the surface of the merry-go-round is 0.300 m/s.

Step 1: Calculate the initial angular momentum of the system

Initially, the merry-go-round is at rest and the boy is not moving relative to it. So, the initial angular momentum of the system (merry-go-round and the boy) is zero.

Step 2: Calculate the final angular momentum of the boy

When the boy starts running, he will have an angular momentum relative to the center of the merry-go-round. The final angular momentum of the boy can be found using the formula: \(L = r \times p\), where \(L\) is the angular momentum, \(r\) is the distance from the center, and \(p\) is the linear momentum of the boy (mass times velocity). $$ L_{boy} = r \times m_{boy} \times v_{boy} $$Plug in the given values: $ L_{boy} = 2.00\,\mathrm{m} \times 25.0\,\mathrm{kg} \times 0.600\,\mathrm{m/s} $$Solve for \(L_{boy}\): $ L_{boy} = 30.0\,\mathrm{kg\,m^{2}/s} $

Step 3: Use conservation of angular momentum

By the conservation of angular momentum principle, the final angular momentum of the system equals the initial angular momentum. As the initial angular momentum was zero, the final angular momentum of the merry-go-round should be equal in magnitude but opposite in direction to the final angular momentum of the boy: $ L_{merry-go-round} = -L_{boy} = -30.0\,\mathrm{kg\,m^{2}/s} $

Step 4: Calculate the angular velocity of the merry-go-round

Now, we can use the relationship between angular momentum and angular velocity to find the angular velocity of the merry-go-round: $ L = I \times \omega \(Where \)L\( is the angular momentum, \)I\( is the moment of inertia, and \)\omega\( is the angular velocity. Rearrange the formula to solve for \)\omega\(: \) \omega = -\frac{L_{merry-go-round}}{I} \(Plug in the given values: \) \omega = -\frac{-30.0\,\mathrm{kg\,m^{2}/s}}{200\,\mathrm{kg\,m^{2}}} \(Solve for \)\omega\(: \) \omega = 0.15\,\mathrm{rad/s} \(So, the angular velocity of the merry-go-round is \)0.15\,\mathrm{rad/s}$.

Step 5: Find the relative speed of the boy on the merry-go-round

To find the speed of the boy relative to the surface of the merry-go-round, we have to find the difference in their velocities. First, find the linear velocity of the merry-go-round surface relative to the center: $ v_{merry-go-round} = r \times \omega $ Plug in the given values: $ v_{merry-go-round} = 2.00\,\mathrm{m} \times 0.15\,\mathrm{rad/s} $ Calculate the linear velocity of the merry-go-round surface relative to the center: $ v_{merry-go-round} = 0.30\,\mathrm{m/s} $ Now, we can find the relative speed of the boy with respect to the merry-go-round surface by subtracting the linear velocity of the merry-go-round surface from the speed of the boy relative to the ground: $ v_{relative} = v_{boy} - v_{merry-go-round} $ Substitute the values: $ v_{relative} = 0.600\,\mathrm{m/s} - 0.30\,\mathrm{m/s} $ Calculate the relative speed: $ v_{relative} = 0.300\,\mathrm{m/s} $ So, the speed of the boy relative to the surface of the merry-go-round is \(0.300\,\mathrm{m/s}\).

Key Concepts

Angular Velocity

Angular velocity is a measure of how fast an object rotates or spins around a particular axis. In this exercise, the merry-go-round starts from rest and spins in response to the boy running around it. The concept of conservation of angular momentum plays a crucial role here. According to this principle, if no external torques are acting on a system, the total angular momentum remains constant.
When the boy begins running, he transfers angular momentum to the merry-go-round. We can calculate the angular velocity (\( \omega \) ) of the merry-go-round using the relation between angular momentum (\( L \) ) and angular velocity:
- \( L = I \times \omega \) This formula shows that angular momentum (\( L \) ) is the product of the moment of inertia (\( I \) ) and the angular velocity (\( \omega \) ).By rearranging the formula, \( \omega \) can be found by dividing the angular momentum by the moment of inertia:
- \( \omega = \frac{L}{I} \) For the merry-go-round, the angular velocity is calculated to be \( 0.15 \, \mathrm{rad/s} \).
This means every second, the merry-go-round rotates by \( 0.15 \, \mathrm{radians} \). It might seem small, but with conserved momentum, it precisely balances the boy's motion.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes to its rotation, similar to how mass measures an object's resistance to changes in its motion along a straight line. In essence, it's how much "inertia" an object has when rotating.
The moment of inertia of an object depends mainly on its mass and how that mass is distributed in relation to the axis of rotation. In our problem, the merry-go-round has a moment of inertia of \( 200 \, \mathrm{kg} \, \mathrm{m}^2 \). This value captures the distribution of the merry-go-round's mass with respect to its center.
The larger the moment of inertia, the harder it is to change the object's rotational speed. That's why even though the boy is sprinting, the merry-go-round rotates at \( 0.15 \, \mathrm{rad/s} \), maintaining balance with the boy's angular momentum.
Below are key points about moment of inertia:- It is expressed in units of \( \mathrm{kg\,m^2} \).- For complex shapes, it is determined by integrating the mass distribution.- Higher values indicate a greater resistance to rotational changes.
Understanding moment of inertia helps explain why objects with more mass spread out rotate slower than more compact ones.

Relative Velocity

Relative velocity is the velocity of an object as observed from a particular frame of reference. In many problems, determining relative velocity is crucial to understand the motion of objects from different perspectives. In this case, we are interested in how fast the boy is moving relative to the surface of the merry-go-round, not the ground.
Using the concept of relative velocity, we find the boy's speed in relation to the spinning surface he's running on. Initially, the boy is running with a speed of \( 0.600 \, \mathrm{m/s} \) relative to the ground. However, as the merry-go-round itself moves, its surface also has a velocity.
Key points when calculating relative velocity:- Identify both velocities: boy's speed relative to the ground and merry-go-round's surface speed.- The surface speed is found by multiplying the angular velocity (\( \omega \) ) with the radius (\( r \) ). \( v_{\text{merry-go-round}} = r \times \omega \) = \( 2.00 \, \mathrm{m} \times 0.15 \, \mathrm{rad/s} = 0.30 \, \mathrm{m/s} \)- The boy’s speed relative to the merry-go-round is \( v_{\text{relative}} = v_{\text{boy}} - v_{\text{merry-go-round}} \).
In this exercise, the resulting speed relative to the merry-go-round is \( 0.300 \, \mathrm{m/s} \).
This calculation highlights that while the boy runs at a certain speed relative to the ground, his effective speed on the spinning platform considers both his motion and the platform's movement.