Question

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 \(g\).

Short answer

a) The three forces acting on the rod include the gravitational force acting downward at the center of mass of the rod (mg), the normal force exerted by the table acting upward at the point of contact with the table (N), and the static friction acting at the point of contact with the table (fs). b) The angular speed of the rod when it makes a 45° angle with respect to the vertical is approximately 2.84 rad/s. The vertical acceleration of the moving end of the rod is approximately 2.02 m/s². The normal force exerted by the table is approximately 1.61 N. c) 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).

Step-by-step solution

Answer to part a:

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 - \(f_s\).

Answer to part b (calculation of angular speed):

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: \(\omega = \sqrt{\frac{12 \times 9.81 \times 0.50 (1 - \cos 45^{\circ})}{0.50^2}} \approx 2.84 \text {rad/s}\)

Answer to part b (calculation of vertical acceleration):

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

Answer to part b (calculation of normal force exerted by the table):

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: \(N = \frac{2}{3}Mg\cos\theta + \frac{1}{2}Mg\left(\frac{d\omega^2}{d\theta}\right) \approx 1.61 \text{ N}\)

Answer to part c (calculation of linear acceleration):

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 (\(g\)).

Key concepts

Rotational Motion

Rotational motion refers to the spinning or turning around of an object about an axis. In this exercise, the uniform rod undergoes rotational motion as it pivots around the point where it touches the table.

• The pivot point acts as a fixed axis, around which the rod rotates as it falls.
• While falling, different parts of the rod move with different linear speeds, but they all share the same angular speed about the pivot point.

Rotational motion is often analyzed using parameters like angular displacement, angular velocity (\( \omega \)), and angular acceleration (\( \alpha \)).In rotational dynamics, knowing how forces and torques act on an object allows us to understand how it rotates. In this problem, the gravitational force causes a torque that leads to the rod's rotational motion.

Conservation of Energy

The conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another.

• In this scenario, gravitational potential energy (when the rod is vertical) is converted into rotational kinetic energy as the rod falls.
• The initial potential energy depends on the rod's position, being high when the rod is upright.
• As the rod falls to the horizontal, this energy is converted into the kinetic energy of rotation.

By calculating the conversion of potential energy to kinetic energy, we can find the angular speed of the rod. The formula used to express this energy transformation in this problem is:\[Mgh = \frac{1}{2}I\omega^2\]This showcases how energy principles can greatly assist in solving mechanical problems that involve rotation.

Angular Speed

Angular speed, also known as angular velocity, is a measure of how quickly an object rotates. It defines the rate of change of the angular position of the object concerning time.

• It is denoted by the Greek letter omega (\( \omega \)), and is measured in radians per second (rad/s).
• As the rod transitions from vertical to horizontal, angular speed provides insight into how fast this motion is occurring.
• The calculation in this exercise uses energy conservation principles to find \( \omega \).

In our situation:\[\omega = \sqrt{\frac{12gh(1-\cos\theta)}{L^2}}\]Here, this expression implies that angular speed depends not only on gravitational acceleration (\( g \)) and the length of the rod (\( L \)), but also the angle (\( \theta \)) it makes with the vertical. This formula highlights that the energy of the system intricately influences the motion.

Torque

Torque is a measure of how much a force acting on an object causes that object to rotate. In the case of the rod falling, torque is critical in understanding its rotational dynamics.

• Torque (\( \tau \)) depends on the force applied, the lever arm distance, and the angle of application.
• The gravitational force acting on the rod's center of mass creates a torque that causes rotation around the pivot point.
• The net torque on the rod determines its angular acceleration, found using \( \tau = I\alpha \).

The magnitude of the torque due to gravity is calculated as:\[\tau = \frac{1}{2}Lmg\sin{\theta}\]This expression indicates that the torque increases with the sine of the angle, which impacts how fast or slow the rod will rotate. Understanding torque allows resolving the forces in the system and illustrating how they govern the rod's rotational behavior.