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

Question: Based on the given solution, identify the forces acting on the rod, and calculate the angular speed, vertical acceleration, normal force when the rod is at a 45-degree angle, and linear acceleration of the end of the rod when it hits the table without slipping. Compare it to the gravitational acceleration g.

Step-by-step solution

Forces acting on the rod

The forces acting on the rod are: 1. Gravitational force (mg) acting on the center of mass of the rod (middle point) 2. Normal force (N) exerted by the table on the rod at its lower end. b) Calculate angular speed, vertical acceleration, and normal force at a 45-degree angle

Moment of inertia

First, we need to find the moment of inertia (I) of the rod about its end. For a uniform rod, the moment of inertia about one end is given by: \(I = \frac{1}{3}mL^2\) Where m is the mass of the rod and L is its length.

Angular acceleration

Next, using Newton's second law for rotational dynamics, we have: \(\tau = I \alpha\) Where \(\tau\) is the net torque acting on the rod about its lower end and \(\alpha\) is the angular acceleration. The net torque is due to the gravitational force (mg) acting at the center of mass of the rod. Hence, the torque about the lower end is given by: \(\tau = \frac{1}{2}mgL\sin\theta\) Now, we can find the angular acceleration: \(\alpha = \frac{\tau}{I} = \frac{3mg\sin\theta}{2mL} = \frac{3g\sin\theta}{2L}\)

Angular speed

Now, we can use the kinematic equation to find the angular speed of the rod when it makes an angle of 45-degrees with the vertical: \(\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)\) Since the rod is released from rest, \(\omega_0 = 0\) and \(\theta_0 = 0\). Therefore, we can find \(\omega\): \(\omega = \sqrt{2\alpha\theta} = \sqrt{\frac{3g\sin\theta}{L}(45^\circ)}\)

Vertical acceleration

We can now find the vertical acceleration (a_y) of the moving end of the rod using: \(a_y = \alpha L\sin\theta\)

Normal force

The normal force (N) exerted by the table on the rod can be found using Newton's second law for linear dynamics in the y-direction: \(N - mg\cos\theta = ma_y\) Solving for N: \(N = mg\cos\theta + ma_y\) c) Calculate linear acceleration of the end point of the rod when it hits the table

Linear acceleration of end point

If the rod falls onto the table without slipping, the linear acceleration (a_x) of the endpoint of the rod in the horizontal direction can be found using: \(a_x = \alpha L\cos\theta\) Now, we can compare it to the gravitational acceleration g. Note: Remember to use the given values for M, L, and theta to find numerical values for the variables calculated above.

Key concepts

Moment of Inertia

Moment of inertia, represented with the symbol 'I', is a measure of an object's resistance to any change in its state of rotation. The formula changes based on the shape of the object and the axis about which it is rotating. For our uniform rod rotating about one end, the moment of inertia is given by the relation:
\[I = \frac{1}{3}mL^2\]
where 'm' stands for the mass of the rod, and 'L' is its length. This calculation is crucial because it tells us how difficult it is to start or stop the rod spinning about one end. It's akin to mass in linear motion—the higher the moment of inertia, the more torque is needed to achieve the same angular acceleration.

Angular Acceleration

Angular acceleration, noted as \(\alpha\), is the rate of change of angular speed over time. In the context of our falling rod, angular acceleration occurs due to the unbalanced torque caused by gravity acting on the rod's center of mass. With the formula
\[\alpha = \frac{\tau}{I}\]
we establish a direct relationship between torque (\(\tau\)), moment of inertia (I), and angular acceleration (\(\alpha\)). As the rod falls, the component of gravitational force perpendicular to the rod generates a torque that causes angular acceleration. The greater the torque or the smaller the moment of inertia, the higher this acceleration will be.

Torque

Torque is a measure of how much a force acting on an object causes that object to rotate. The amount of torque depends on the force applied, the distance from the pivot point (or axis of rotation), and the angle of the force. For the falling rod, the formula is
\[\tau = \frac{1}{2}mgL\sin(\theta)\]
In this situation, gravity pulls the rod down at its center of mass, creating a torque around the point of contact with the table because there's a horizontal 'lever arm' involved. The \(\sin(\theta)\) factor adjusts the torque value based on the rod's angle relative to the vertical, with maximum torque when the rod is horizontal.

Angular Speed

Angular speed, symbolized as \(\omega\), is the rate at which an object rotates around an axis, measured in radians per second. We used the kinematic equation \(\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)\) to determine the angular velocity of our rod at the 45-degree angle. This equation relates the initial angular speed (\(\omega_0\)), angular acceleration (\(\alpha\)), and the change in angle (\(\theta - \theta_0\)). Since the rod begins from rest, the initial angular speed is zero, which simplifies our calculation.

Normal Force

Normal force is the perpendicular force exerted by a surface to support the weight of an object. It’s a reactive force that responds to other forces placed on a body in contact with a surface. For the rod, as it falls and makes an angle with the vertical, the table exerts an upward normal force (N) at the lower end. The calculation integrates the vertical components of other forces, including the rod’s weight and the force due to angular acceleration, to find the total normal force exerted by the table using Newton's second law.

Linear Acceleration

Linear acceleration is the rate of change of velocity with time in a straight line. While we’ve been focusing on the rod’s rotation, it’s also important to consider the linear acceleration of the end point as it hits the table. This value gives us a sense of how the translational motion of the rod changes. The end of the rod accelerates horizontally with \(a_x = \alpha L\cos(\theta)\) and vertically with \(a_y = \alpha L\sin(\theta)\), showing that linear acceleration is directly related to the angular acceleration and the length of the rod.