Problem 60 - University Physics with Modern Physics

Question

A circular platform of radius \(R_{p}=4.00 \mathrm{~m}\) and mass \(M_{\mathrm{p}}=400 .\) kg rotates on frictionless air bearings about its vertical axis at 6.00 rpm. An 80.0 -kg man standing at the very center of the platform starts walking \((\) at \(t=0)\) radially outward at a speed of \(0.500 \mathrm{~m} / \mathrm{s}\) with respect to the platform. Approximating the man by a vertical cylinder of radius \(R_{\mathrm{m}}=0.200 \mathrm{~m}\) determine an equation (specific expression) for the angular velocity of the platform as a function of time. What is the angular velocity when the man reaches the edge of the platform?

Step-by-Step Solution

Answer

**Answer**: The angular velocity of the platform when the man reaches the edge is \(\omega_p = \frac{4}{5}\frac{\pi}{5} \, rad/s\).

Step 1: Understand the given values

In this problem, we are given: - Radius of the platform: \(R_p = 4.00 \, m\) - Mass of the platform: \(M_p = 400 \, kg\) - Platform rotates with initial angular velocity: \(6.00 \, rpm\) - Mass of the man: \(M_m = 80.0 \, kg\) - Radial speed of the man: \(v_r = 0.500 \, m/s\)

Step 2: Conservation of angular momentum

The total angular momentum of the system is conserved because there are no external torques acting on it. The angular momentum of the platform (\(L_p\)) is given by \(L_p = I_p \omega_p\), where \(I_p\) is the moment of inertia of the platform and \(\omega_p\) is its angular velocity. Similarly, the angular momentum of the man (\(L_m\)) is given by \(L_m = I_m \omega_m\), where \(I_m\) is the moment of inertia of the man and \(\omega_m\) is his angular velocity. Since the angular momentum is conserved, we have: \(L_{total} = L_p + L_m = I_p \omega_p + I_m \omega_m\)

Step 3: Calculate the initial angular velocity in rad/s

To find the initial angular velocity in radians per second, we can use the conversion factor: \(\omega_{initial} = 6.00 \,rpm \times \frac{2\pi}{60} = 2\pi \, rad/min \times \frac{1}{60} = \frac{\pi}{5} \, rad/s\)

Step 4: Calculate the moment of inertia for the platform and the man

To compute the moment of inertia of the platform, we can treat it as a solid disk. The formula for the moment of inertia for a solid disk is \(I_p = \frac{1}{2} M_p R_p^2\). For the man, we can approximate him as a cylinder, so the formula is \(I_m = \frac{1}{2} M_m r^2\), where \(r\) is the distance between the center of the platform and the man at any given time. \(I_p = \frac{1}{2}(400)(4^2) = 3200 \, kg \cdot m^2\) \(I_m = \frac{1}{2}(80)r^2\)

Step 5: Equation for the angular velocity of the platform

Using the conservation of angular momentum equation from step 2, we can find an equation for the platform's angular velocity as a function of time: \(L_{total} = I_p\omega_p + I_m\omega_m\) We know that \(L_{total}\) is constant because the total angular momentum is conserved. As the man moves outward, his moment of inertia increases while the platform's moment of inertia decreases. Therefore, the man's angular velocity \(\omega_m\) will be equal to \(\omega_p - v_r / r\). So, we can rewrite the equation as: \(I_p\omega_p + I_m(\omega_p - v_r / r) = L_{total}\) Now, we have an equation relating \(\omega_p\), \(r\), and \(v_r\). We can solve for \(\omega_p\) as a function of time by considering the position of the man, \(r(t) = v_r t\). Plug it into the equation above: \(I_p\omega_p + I_m(\omega_p - o.5 / (0.5t)) = L_{total}\) Rearrange and solve for \(\omega_p(t)\): \(\omega_p(t) = \frac{L_{total}}{I_p + I_m(t)}\)

Step 6: Calculate the angular velocity when the man reaches the edge

When the man reaches the edge of the platform, \(r = R_p = 4\,m\). So, we have: \(I_m = \frac{1}{2}(80)(4)^2 = 640\, kg\cdot m^2\). Now, we can plug this value into the equation for \(\omega_p(t)\) and calculate the final angular velocity: \(\omega_p(t) = \frac{L_{total}}{3200 + 640} = \frac{L_{total}}{3840}\) We know the initial angular momentum is \(L_{total} = I_p \omega_{initial} = 3200\frac{\pi}{5}\). Plug it into the equation: \(\omega_p(t) = \frac{3200 \pi/5}{3840} = \frac{4}{5}\frac{\pi}{5}\) Therefore, the angular velocity of the platform when the man reaches the edge is \(\omega_p = \frac{4}{5}\frac{\pi}{5} \, rad/s\).

Key Concepts

Angular Velocity

Angular velocity is a measure of how fast an object rotates around a particular axis. It is usually expressed in radians per second (rad/s).
In the context of our exercise, the platform begins its rotation with an initial angular velocity of 6.00 revolutions per minute (rpm), which we convert into radians per second for easier calculations. Remember, one complete revolution is equivalent to a full circle or 2π radians. To convert from rpm to rad/s, we use the formula:

• \( ext{Angular velocity in rad/s} = ext{Angular velocity in rpm} \times \frac{2\pi}{60} \)

This gives us the angular velocity in rad/s, facilitating further computation with integration into the conservation of angular momentum equation. Importantly, as the man walks from the center towards the edge of the platform, the total angular velocity of the system changes. This is calculated to ensure that the conservation of angular momentum is maintained, which also impacts the rotational speed of the platform. By formulating the platform's angular velocity \( \omega_p(t) \) as a function of time, we understand how the system dynamically adjusts to ensure equilibrium.

Moment of Inertia

The moment of inertia is like the rotational equivalent of mass in linear motion. It determines how much torque is needed for an object to achieve angular acceleration. For a rotating body, the moment of inertia depends on both its mass and the distribution of the mass around the axis of rotation.
In this exercise, the platform is treated as a solid disk, while the man is approximated as a cylinder.

• The moment of inertia for the platform \( I_p = \frac{1}{2} M_p R_p^2 \)
• The moment of inertia for the man \( I_m = \frac{1}{2} M_m r^2 \), where \( r \) is the distance from the center.

Understanding how the moment of inertia works helps us in determining the angular velocity. When the man moves radially outward, his contribution to the system's overall moment of inertia changes. This owes to the increased distance from the rotation axis, thus increasing the system's resistance to change in its rotational state. Consequently, the platform's rotational speed (angular velocity) changes to conserve angular momentum.

Radial Motion

Radial motion refers to movement towards or away from a center point, in this case, the center of the rotating platform. It directly affects the distribution of mass and the moment of inertia in a rotating system.
For our exercise, the man walks radially outward from the center of the platform at a speed of \( 0.500 \ m/s \). As the man moves away from the center, his distance \( r(t) \) from the center increases with time \( t \).

• \( r(t) = v_r \cdot t \), where \( v_r = 0.500 \ m/s \)

This radial movement is crucial because as the distance \( r \) increases, so does the man's moment of inertia \( I_m = \frac{1}{2} M_m r^2 \). Being inversely proportional to the angular velocity, an increase in \( r \) will decrease the angular speed (if no other changes occur), ensuring that angular momentum is conserved. Thus, radial motion is a critical factor influencing how rotational dynamics behave and are a core aspect of understanding system stability and angular momentum conservation.