Question

A circular platform of radius \(R_{p}=4.00 \mathrm{~m}\) and mass \(M_{\mathrm{p}}=400 . \mathrm{kg}\) rotates on frictionless air bearings about its vertical axis at 6.00 rpm. An \(80.0-\mathrm{kg}\) 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?

Short answer

Answer: The expression for the angular velocity of the platform as a function of time is ω_final(t) = π / (1 + 0.05t^2). When the man reaches the edge of the platform, the angular velocity is approximately 0.118 rad/s.

Step-by-step solution

Conservation of Angular Momentum

Angular momentum is conserved since there are no external torques acting on the man-platform system. Thus, the initial angular momentum of the system equals the final angular momentum. Let's call the initial angular momentum L_initial and the final angular momentum L_final. So we have: L_initial = L_final

Initial Angular Momentum

Initially, the man is standing at the center of the platform, so his contribution to the angular momentum is 0. The platform's initial angular momentum can be expressed as: L_initial = I_platform * ω_initial where I_platform is the moment of inertia of the platform, and ω_initial is the initial angular velocity of the platform (6.00 rpm, which needs conversion to rad/s). I_platform = M_p * R_p^2 / 2 = 400 * 4^2 / 2 = 3200 kg.m^2 ω_initial = 6.00 * (2 * π / 60) = π rad/s L_initial = 3200 * π kg.m^2/s

Final Angular Momentum

The final angular momentum takes into account both the platform and the man. We have: L_final = I_platform * ω_final + I_man * ω_final where I_man is the man's moment of inertia (which can be approximated as a cylinder) and ω_final is the final angular velocity of the platform. I_man = M_m * r^2 + M_m * (R_m)^2 / 2 Here, r is the distance of the man from the center of the platform. Since the man is walking at a constant speed, we can write r as: r = 0.5t (in meters, where t is in seconds) I_man = 80 * (0.5t)^2 + (80 * 0.2^2 / 2) = 160t^2 kg.m^2 Substituting this into the expression for L_final: L_final = (3200 + 160t^2) * ω_final

Equating Initial and Final Angular Momentum

Using the conservation of angular momentum, we can set L_initial = L_final: 3200 * π = (3200 + 160t^2) * ω_final Dividing both sides by 3200: π = (1 + 0.05t^2) * ω_final Now we can get the expression for angular velocity as a function of time: ω_final(t) = π / (1 + 0.05t^2)

Angular Velocity When Man Reaches the Edge

To find the angular velocity when the man reaches the edge of the platform, we need to find the time it takes for the man to reach the edge. Since the platform has a radius of 4 meters and the man is walking at 0.5 m/s, it will take: t_edge = 4 / 0.5 = 8 seconds Now we can substitute this time into our equation for ω_final: ω_final(8) = π / (1 + 0.05 * 8^2) = π / (1 + 25.6) = π / 26.6 ω_final(8) ≈ 0.118 rad/s Thus, when the man reaches the edge of the platform, the angular velocity of the platform is approximately 0.118 rad/s.

Key concepts

Conservation of Angular Momentum

Understanding the principle of conservation of angular momentum is fundamental in rotational motion problems. According to this principle, if no external torque acts on a system, the total angular momentum of the system remains constant. This is comparable to the conservation of linear momentum, where the total momentum remains constant in the absence of external forces.

In the exercise, a man walks radially outward on a rotating circular platform. The system initially includes just the spinning platform, as the man contributes no angular momentum from the center. The initial angular momentum depends on two things: the moment of inertia of the platform and its initial angular velocity. This relationship is encapsulated by the equation: \[\begin{equation}L_{\text{initial}} = I_{\text{platform}} \times \omega_{\text{initial}}\end{equation}\]When the man walks outwards, both his and the platform's moment of inertia change, which would change the angular momentum if there were no conservation law. To maintain the same total angular momentum (as no external torques act on the system), the platform's angular velocity must decrease. This inverse relationship ensures conservation. By calculating the system's initial and final angular momenta and setting them equal, we resolve the final angular velocity as a function of time.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotation and is dependent on both the object's mass and the distribution of that mass relative to the axis of rotation. The more mass far away from the axis, the greater the moment of inertia. It plays a similar role in rotational motion to that of mass in linear motion.

For the round platform in the exercise, the moment of inertia is determined by the formula for a solid disk: \[\begin{equation}I_{\text{platform}} = \frac{1}{2} M_{\text{p}} R_{\text{p}}^2\end{equation}\]For the man, modeled as a cylinder, the calculation needs to account for both his movement away from the center of the platform and his cylindrical shape. The further he moves, the larger his contribution to the moment of inertia of the system because his mass is further from the axis of rotation:

• At the center, his moment of inertia is minimal.
• As he walks outward, his distance from the axis, and thus his moment of inertia, increases.

This changing moment of inertia for the man as he walks outward directly influences the system's total moment of inertia and thus the angular velocity over time, according to the conservation of angular momentum.

Rotational Kinematics

Finally, let's delve into rotational kinematics, which describes the motion of objects in a rotational context. Rotational kinematics involves angular position, angular velocity, and angular acceleration, analogous to their linear counterparts (position, velocity, and acceleration).

In our example exercise, we are concerned with the angular velocity, which describes the rate of change of angular position over time. Initially, the platform has a constant angular velocity, but this changes when the man walks outward. Angular velocity can be expressed as:\[\begin{equation}\omega = \frac{\theta}{t}\end{equation}\]where \( \theta \) is the angular position and \( t \) is time. The final angular velocity must be solved using the principles of conservation of angular momentum and the changing moment of inertia.

We used a specific expression for the man's distance from the platform's center as a function of time to determine how the system's angular velocity changes. This way, we link the linear motion of the man walking with the rotational motion of the platform, showcasing the interconnectedness of linear and rotational kinematics.