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?

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: \(v_r = 0.500 \, m/s\)

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: \(L_{total} = L_p + L_m = I_p \omega_p + I_m \omega_m\)

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

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. \(I_p = \frac{1}{2}(400)(4^2) = 3200 \, kg \cdot m^2\) \(I_m = \frac{1}{2}(80)r^2\)

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 \(\omega_p(t)\): \(\omega_p(t) = \frac{L_{total}}{I_p + I_m(t)}\)

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 \(\omega_p = \frac{4}{5}\frac{\pi}{5} \, rad/s\).