A beam A B with a mass m and of uniform cross-section is...

Key Concepts

Free Body Diagram Analysis

Drawing a free body diagram is a fundamental step in solving dynamics problems. It involves isolating the beam and representing all the forces and moments acting upon it. This method helps systematically apply Newton’s laws to identify the net torque and force, forming the basis for solving for angular and linear accelerations.

Center of Mass (and Mass Distribution)

The center of mass represents the average location of all the mass in the body and is crucial in analyzing the translational motion of the beam. When a support is lost, the gravitational force acting through the center of mass combines with the reactive force from the remaining spring to produce net translational acceleration.

Kinematics of Rigid Body Motion

This concept connects the angular motion of the body with the linear motion of its points. Once the angular acceleration is known, the linear acceleration at any point on the beam can be determined by considering the tangential and radial components of the motion, providing answers for specific points like A and B.

Newton’s Second Law for Rotation

This law states that the net torque acting on a rigid body is equal to the product of its moment of inertia and its angular acceleration. It is pivotal in determining the rotational response of the beam when one of the springs fails, as it allows the calculation of angular acceleration from the torque produced by the remaining forces.

Rigid Body Dynamics

This concept involves analyzing the motion of a solid body when forces and torques are applied. In problems involving beams or other extended bodies, it is essential to consider both translational and rotational motion, treating the body as a rigid object that does not deform. In this scenario, the sudden loss of one support changes both the net force and net torque, leading to rotation and linear acceleration.

Moment of Inertia

The moment of inertia quantifies how the mass of a body is distributed relative to a chosen axis of rotation and is a measure of the body’s resistance to angular acceleration. For a beam of uniform cross-section, it is a critical value for linking the net applied torque to the resulting angular acceleration.

Transcript

00:01 So here in this problem we are given a beam ab that has a mass m and it has a uniform cross -section and it is suspended by two spring as you can see in this figure now spring to breaks and then we have to determine three things at the instant first one is the angular acceleration of the beam second the acceleration of the point a this point and finally the acceleration of the point b now to solve this problem it will be considering the free body diagram before the spring to breaks so here the force would be acting would be in the upper direction and it would be f let it be f1 and here let it be f2 and the weight would be acting in the vertical downward direction this is the centroid or the center of gravity here m .g would be acting in the vertical downward direction now the figure is symmetric that means the force f1 is equals to the force f2 and equals to m g over 2 now next i will be considering the freeboard diagram of the beam when the spring two breaks so this is a free border diagram of the beam when the spring two breaks and this is a kinetic diagram now here we know f1 is it is m g by 2 and the moment of inertia about the centroid for the beam it is equals to i is equals to emil square divide by 12 now next i will be taking the moment of all the forces about the center of gravity of the beam so sigma m g is equals to i alpha here g is this point at which this is the center of gravity now sigma m g would be in this case it would be this is f1 times this distances from here to here it would be l by 4 so l by 4 and it is equals to m l square divide by 12 this is the value of i the moment of inertia times alpha now from here we're going to solve for alpha and we can put the value of f1 as m g over 2 here so this is m g over 2 times l over 4 is equals to m l square divide by 12 times alpha now here this m and m will cancel out as similarly this l and this l will cancel out and i will get the expression for alpha as that is the angular acceleration of the beam as 3g over 2l so is the answer for the a part now next i will be solving the v part where we have to calculate the acceleration of the point a so the acceleration at point a it will be equals to acceleration at point g minus the relative acceleration at point a with to g next i will be calculating the acceleration at point g for that we can use the neuron second of motion that is sigma f y is equals to mass times acceleration so here we have w minus f1 is equals to mass times acceleration we just apply the neutral signal of motion here so if one we know it is m g over 2 and w it is m g so m g minus m g over 2 is equal to mass time and this is the acceleration at point g so mass times acceleration at point g so from here acceleration at point g is equals to g over 2 and the direction would be in the vertically downward direction so it is equals to ag we have calculated now next we'll be calculating the relative acceleration at point a with respect to g so for that we know that a the relief acceleration with point a with respect to g is equals to l over 2 times alpha this distance was l over 2 from point a to the center of gravity this total distance would be l over 2 so from here l over 2 as it is and alpha we have calculated as this much that is 3g over 2l now we have ag also and this one also we can plug it here to calculate the acceleration at point a so we'll get acceleration at point a a is equals to ag we have g by 2 minus and this direction would be in the upward direction that is why the minus came here so this is 3g over 12 so from here we'll get the final answer as minus g by 4 that means the acceleration at point a would be acting in the upward direction because in the downward direction we have taken acceleration to be positive and the in the upward direction the acceleration would be negative so we can write that acceleration at point a is equal to g over 4 and the direction would be in the vertically upward direction.

05:31 Now actually i did a mistake here it should be 3g over 2 because this l and l will cancel it out.

05:38 Now we'll get the acceleration as minus g over 4...

Please give Ace some feedback