A uniform rod of mass \(M=250.0 \mathrm{~g}\) and length \(L=50.0 \mathrm{~cm}\) stands vertically on a horizontal table. It is released from rest to fall. a) What forces are acting on the rod? b) Calculate the angular speed of the rod, the vertical acceleration of the moving end of the rod, and the normal force exerted by the table on the rod as it makes an angle \(\theta=45.0^{\circ}\) with respect to the vertical. c) If the rod falls onto the table without slipping, find the linear acceleration of the end point of the rod when it hits the table and compare it with \(g\).

There are three forces acting on the rod. These are: 1. The gravitational force acting downward at the center of mass of the rod - \(mg\). 2. The normal force exerted by the table acting upward at the point of contact with the table - \(N\). 3. The static friction acting at the point of contact with the table - \(f_s\).

As the rod falls, it starts rotating about the point of contact with the table. We can use conservation of energy to find the angular speed. Let \(I_{cm}\) be the moment of inertia of the rod about an axis passing through its center of mass and parallel to the table. We have \(I_{cm} = \frac{1}{12}ML^2\). Using the parallel axis theorem, we can find the moment of inertia of the rod about the point of contact with the table: \(I = I_{cm} + MD^2\), where \(D = \frac{L}{2}\), which gives \(I = \frac{1}{3}ML^2\). When the rod makes an angle \(\theta\) with respect to the vertical, the gravitational potential energy of the center of mass decreases by an amount \(U = Mgh\), where \(h = \frac{L}{2}\left(1-\cos\theta\right)\). As the rod is in pure rotation, it has rotational kinetic energy: \(K = \frac{1}{2}I\omega^2\). Applying conservation of energy, we will have: \(U = K \Rightarrow Mgh = \frac{1}{2}I\omega^2\) Solving for the angular velocity \(\omega\), we get: \(\omega^2 = \frac{6Mgh}{ML^2} = \frac{12gh}{L^2}\) \(\omega = \sqrt{\frac{12gh}{L^2}}\) We substitute the given values for \(g = 9.81 \text{ m/s}^2\), \(L=0.50 \text{ m}\), and \(\theta = 45^{\circ}\) to find the angular speed: \(\omega = \sqrt{\frac{12 \times 9.81 \times 0.50 (1 - \cos 45^{\circ})}{0.50^2}} \approx 2.84 \text {rad/s}\)

To find the vertical acceleration of the moving end of the rod, we can use the formula for centripetal acceleration: \(a = \omega^2r\). Here, \(r = L\): \(a = \omega^2L = \left(2.84 \text{ rad/s}\right)^2 \times 0.50 \text{ m} \approx 2.02 \text{ m/s}^2\)

We can use the equation for torque about the point of contact with the table to find the normal force N: \(\tau = I\alpha\) Here, \(\alpha\) is the angular acceleration of the rod. To find the angular acceleration, we need to differentiate the equation for the angular velocity \(\omega\): \(\frac{d\omega^2}{dt^2} = \frac{2 \omega d\omega}{dt} = 2\alpha\) \(\alpha = \frac{d\omega^2}{d\theta}\frac{d\theta}{dt} = \omega\frac{d\omega^2}{d\theta}\) The torque produced by gravity is due to the gravitational force acting at the center of mass of the rod at a distance of \(L/2\) from the pivot point: \(\tau = -\frac{1}{2}Lmg\sin{\theta}\) Equating the torques, we get: \(-\frac{1}{2}Lmg\sin{\theta} = \frac{1}{3}ML^2\alpha\) Here, \(\alpha = \omega\frac{d\omega^2}{d\theta}\) and \(\omega^2 = \frac{12gh(1-\cos\theta)}{L^2}\). Solving the above equation for \(N\) and substituting the given values, we get: \(N = \frac{2}{3}Mg\cos\theta + \frac{1}{2}Mg\left(\frac{d\omega^2}{d\theta}\right) \approx 1.61 \text{ N}\)

When the rod falls onto the table without slipping, the linear acceleration of its end point is related to its angular velocity: \(a_{linear} = r\alpha = L\alpha\) Using the equation for \(\alpha\) from part b, we get: \(a_{linear} = L\omega\frac{d\omega^2}{d\theta}\) Substitute the values of \(g\), \(L\), and \(\theta = 0\) (as it hits the table) into the equation, we get: \(a_{linear} = 2 \sqrt{3}g \approx 1.20g\) This shows that the linear acceleration of the end point of the rod when it hits the table is \(1.20g\), which is larger than the acceleration due to gravity (\(g\)).