A uniform slender bar A B of mass m is suspended as shown from a uniform disk of the same mass m

A uniform slender bar A B of mass m is suspended as shown...

Key Concepts

Relative Motion in Connected Systems

In systems where components are connected or suspended, constraints influence the motion of each part. Analyzing the relative motion and acceleration of different points requires an understanding of how the translational and rotational motions are coupled through these constraints. This is particularly useful when forces lead to both rotation about a point and translation of the overall mass.

Moment of Inertia and Angular Acceleration

The moment of inertia quantifies a body's resistance to angular acceleration about a given axis. In problems involving rotational motion, knowing the moment of inertia is crucial for relating the applied torque to the resulting angular acceleration. This concept is integral when the body rotates about a pivot or its center of mass after a force is applied.

Rigid Body Kinematics

Rigid body kinematics deals with how different points on a rigid body move relative to each other. It provides the relationship between the linear acceleration of any point on the body, the acceleration of a reference point, and the angular acceleration about that reference point. This concept is essential when determining the accelerations at various points along a body that is both translating and rotating.

Newton's Second Law (Translation and Rotation)

This concept applies Newton's second law to both the translational motion of the center of mass and the rotational motion about the center of mass or a pivot point. For a rigid body, external net forces produce an acceleration of the center of mass, while external torques produce angular acceleration. Both aspects must be considered when analyzing the motion of bodies subjected to forces.

Transcript

00:01 Hello friends as shown in the figure slater bar a b of mass m it's suspended as shown from a uniform disk of same mass so mass of disc is also m there is no friction we have to calculate acceleration of points a and b immediately after a horizontal force p is applied at b just after p is applied let us start solving it kinematics of the system we will analyze kinematics one by one first for cylinder then for rod this is the cylinder a it having acceleration towards right this is point of contact c and angular acceleration alpha a omega is zero motion is rolling without sliding so acceleration of c in horizontal direction is zero acceleration of a is towards right with this positive so it will be acx plus acceleration of a with respect to c so zero plus r alpha a so acceleration of a you will get r alpha a right now for rod a v this is a v rod here acceleration is a this is center of mass having acceleration a a v and angular acceleration alpha a v this is b point this distance would be l y2 and l y2 because this would be also geometrical center of it angular acceleration of the rod will be acceleration of a upon r in clockwise direction acceleration of a along right will be ly2 alpha av minus acceleration of center of mass so you can write acceleration of a in vector form, say equation by now drawing free body diagram fvd of cylinder.

05:20 This is the reaction at a a x and a y.

05:32 This is c point here.

05:34 It is normal reaction and this will be the friction.

05:47 It having mass into acceleration of a it will be i into alpha a v calculating moment of couple about c point and clockwise direction taking to be positive it will be a x into r to be equal to m a into r plus moment of inertia into angular acceleration so we can here write a x into r mass of acceleration a into our moment of energy of the disk about the center of mass mr square into acceleration of a upon r so solving this equation for a x you will get reaction at x will be 3 by 2 m into a and from equation 1 acceleration of a is ly 2 alpha av minus a bar a v say equation number 2 now free body diagram of rod at a reaction are a y and a x weight of the rod its geometrical center force at this is p this is geometrical center so it will be ly 2 and this is also ly 2 and center of mass is moving towards left with mass into acceleration of av.

09:08 Calculating moment of force about a point we can write p into l to be mass into acceleration of av into ly2 plus moment of inertia into angular acceleration of a v.

09:44 Substituting the value of moment of inertia moment of inertia moment of units of the rod about the center of masses 1 by 12 m l square alpha a b say equation number 3 if we take summation of forces towards left to be positive then we can write p plus a x to be mass into acceleration of center of mass substituting from equation 2 we can write p plus 3x2m ly2a v a v minus acceleration of center of mass of the rod a v is equal to mass into acceleration of a v so from here we can write p is equal to 5 by 2 mass of acceleration of center of the rod a v minus 3 by 4 m l angular acceleration of a v say equation number 4 by solving equation 4 and 3 we can find acceleration of rod a v about the center of mass will be 10 by 7 p upon m say equation number 5 now by equation 5 and 3 p can be written as 5 by 2 into m into 10 p upon 7m minus 3 by 4 m l alpha a v so p you will get by solving this equation can be written as minus 18 by 7 p is called to 3 by 4 m l alpha a v so by this equation.

13:07 So by this equation is angular acceleration of rod av can be written as 24 by 7 p upon m into l in clock by direction...

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