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- goround.

Short answer

Question: Calculate the angular velocity of a frictionless merry-go-round as a boy starts to run with a given speed. Then, find the speed of the boy relative to the surface of the merry-go-round. Answer: The angular velocity of the merry-go-round is 0.150 s\(^{-1}\). The speed of the boy relative to the surface of the merry-go-round is 0.300 m/s.

Step-by-step solution

Determine the initial angular momentum of the system

Initially, both the boy and the merry-go-round are at rest which means their initial angular momentum is zero. So, the total initial angular momentum (L_initial) of the system is zero: L_initial = 0

Determine the final angular momentum of the system

When the boy starts running, he has a linear momentum of: \(m_bv_b = 25.0 \mathrm{~kg} \times 0.600 \mathrm{~m/s} = 15.0 \mathrm{~kg\cdot m/s}\) where \(m_b\) is the boy's mass and \(v_b\) is his linear speed relative to the ground. As he's running in a circle with a radius of 2.00 m, the boy's angular momentum can be calculated as: \(L_b = m_b v_b r = 15.0 \mathrm{~kg\cdot m/s} \times 2.00 \mathrm{~m} = 30.0 \mathrm{~kg\cdot m^2/s}\) where \(r\) is the distance from the center of the merry-go-round. However, since the initial angular momentum was zero, the merry-go-round will also have an angular momentum \(L_m = -L_b = -30.0 \mathrm{~kg\cdot m^2/s}\) in the opposite direction to conserve angular momentum.

Calculate the angular velocity of the merry-go-round

With the moment of inertia of the merry-go-round (\(I_m\)) provided, we can now find its angular velocity (\(\omega_m\)) using the formula: \(\omega_m = \frac{-L_m}{I_m}\) \(\omega_m = \frac{30.0 \mathrm{~kg\cdot m^2/s}}{200 \mathrm{~kg\cdot m^2}} = 0.150 \mathrm{~s}^{-1}\) So, the angular velocity of the merry-go-round is 0.150 s\(^{-1}\). Now, let's move to part b:

Calculate the tangential velocity of the merry-go-round

Now that we know the merry-go-round's angular velocity, we can find its tangential velocity (\(v_m\)) at the point where the boy is running: \(v_m = r \omega_m = 2.00 \mathrm{~m} \times 0.150 \mathrm{~s}^{-1}= 0.300 \mathrm{~m/s}\)

Calculate the relative speed of the boy

We will now calculate the speed of the boy relative to the merry-go-round's surface (\(v_{b-m}\)) using vector subtraction: \(v_{b-m} = v_b - v_m\) \(v_{b-m} = 0.600 \mathrm{~m/s} - 0.300 \mathrm{~m/s} = 0.300 \mathrm{~m/s}\) So, the speed of the boy relative to the surface of the merry-go-round is 0.300 m/s.

Key concepts

Angular Velocity

Angular velocity is a measure of how fast an object rotates or revolves around a fixed point or axis. In the context of our exercise, the angular velocity (\( \boldsymbol{\text{ω}} \) ) of the merry-go-round describes how many radians it turns through per second.To understand why radians are used, imagine the merry-go-round making a full rotation. This rotation sweeps out a circular arc with an angle measuring 2π radians. If the merry-go-round completes this rotation over a certain period, we would divide 2π by that time to get its angular velocity. The relevance of angular velocity becomes apparent when conserving angular momentum; since the boy and the merry-go-round system must conserve angular momentum, a change in the boy's motion (like starting to run) necessitates a corresponding change in the angular velocity of the merry-go-round.Calculating angular velocity involves the moment of inertia and the final angular momentum. As our exercise demonstrates, once we know the final angular momentum induced by the boy's movement, we can deduce the merry-go-round's angular velocity using the moment of inertia of the system. It's crucial to understand that angular momentum and angular velocity are directly related - if one changes, the other must change as well if the moment of inertia remains constant.

Moment of Inertia

The moment of inertia (\( \boldsymbol{I} \) ) is akin to mass in linear motion, but it applies to rotational motion. It quantifies an object's resistance to changes in its rotational speed. You can think of it as 'rotational inertia'. Each particle of the object contributes to the moment of inertia; the further the particle is from the axis of rotation, the more it contributes.In our example, the merry-go-round has a sizeable moment of inertia, representing how difficult it is to change its rotation speed. It's affected by both the mass distribution and the shape of the object. For the merry-go-round, which is likely circular and has mass evenly distributed, the value for the moment of inertia is provided. However, calculating the moment of inertia for other shapes involves integrating over the object's volume, considering the distribution of mass relative to the axis of rotation.

Why It Matters

When we analyze our system trying to conserve angular momentum, the moment of inertia is the rotational equivalent to mass in Newton's second law (\( \boldsymbol{F=ma} \) ). Hence, when the boy starts running and begins rotating around the merry-go-round, his motion impacts the rotational speed of the system, which is wholly dependent on the system's moment of inertia.

Tangential Velocity

Tangential velocity (\( \boldsymbol{v} \) ) is the linear velocity of an object moving along a circular path. It's the speed at which a point on the circumference of the rotation is travelling, always directed perpendicular to the radius at that point – hence 'tangential'. It's directly proportional to the rotational speed (angular velocity) and the radius of the path the object is following.To connect this to our example, once the merry-go-round's angular velocity is known, we can calculate the tangential velocity at any point on its edge. The closer to the center, the slower the tangential velocity; the farther from the center, the faster. This change happens because while all points on a rotating body have the same angular velocity, their tangential velocities differ due to their varying distances from the center.

Practical Implication

In the problem, we are asked to find the boy's speed relative to the merry-go-round. We must consider the tangential velocity because although the boy is moving linearly relative to the ground, he is moving along a tangential path relative to the merry-go-round. Hence, we need to calculate the tangential velocity of the merry-go-round's surface at the boy's location to get the boy's relative speed. Such a concept is not only essential in physics problems but also in understanding real-world scenarios, like the take-off speed needed for an airplane considering the Earth's rotation.