A 4-lb bar is attached to a 10 -lb uniform cylinder by a...

Key Concepts

Free Body Diagram

A free body diagram (FBD) is a simplified representation of a system or mechanical component isolated from its surroundings, showing all external forces and moments acting on it. It is crucial for setting up the equations of equilibrium or motion, as it clearly depicts how forces interact at connection points such as pins.

Rolling Without Slipping

Rolling without slipping is a kinematic condition where the point of contact between a rolling object and the surface remains momentarily at rest relative to the surface. This constraint connects the translational and rotational motions of the object, enabling the use of geometrical and kinematic relationships in the analysis of dynamic systems.

Rotational Dynamics

Rotational dynamics involves the study of how torques affect the angular motion of a body. It incorporates concepts such as moment of inertia, angular acceleration, and torque, which are essential when analyzing systems where components rotate or are connected through pin joints.

Equilibrium of Forces and Moments

The principle of equilibrium requires that the sum of all forces and the sum of all moments (torques) acting on a system be zero for a static situation, or balanced with inertial forces in dynamic cases. Establishing these equilibrium equations is key to solving for unknown reaction forces at joints or connection points in mechanical structures.

Transcript

00:01 Hello friends as shown in the figure there is a war of weight 4 pounds and of length 20 inch is attached to the cylinder of weight 10 pounds and radius 16 inch value of edge is given 8 inch and angle theta is 20 degree if cylinder is rotating in clock by direction with angular velocity of two radian per second, calculate the reaction at p.

01:18 So this is the point p at which we have to calculate the reaction.

01:22 First we will calculate moment of inertia of the bar and that of cylinder.

01:33 The mass of the bar is weight divided by acceleration.

01:40 To gravity.

01:42 Weight is given 4 pounds and acceleration due to gravity is 32 .2.

01:48 So this weight becomes point 1 to 4 double to slug and its moment of inertia about center of mass is 1 by 12th of product of mass of the rod and square of its length.

02:10 Substituting the value, its mass is 0 .12 length is 20 inch we are using it in fit so divided by 12 so on solving it moment of inertia about the center of mass you will get 0 .28775 slug into fit square mass of the cylinder weight of cylinder divided by acceleration due to gravity weight is given 20 pounds sorry it is given 10 32 .2 so it would be 0 .31 056 luxe moment of inertia of the cylinder about its center of mass half of mc into r square mass is 0 .31056 and radius r is 16 .12126 and radius r is 16 upon 12 so you will get 0 .27605 luk and 2 fit square now we have to discuss the kinematics of the cylinder and the rod p point g at p point acceleration is tangential and normal center of mass ag this is instantaneous center c having acceleration is c normal alpha let us see here acceleration angular acceleration is alpha in clockwise direction so acceleration of center of mass ag add in the direction 20 degree down to the horizontal for rolling without sylphi that is for pure rolling ac will be 0 and ac is defined as ag minus r alpha that must be 0 so ag is called to r alpha acceleration of the in tangential direction acceleration of g plus h into alpha so you may write ag is r into alpha acceleration of p point in normal direction 0 plus omega is now let us see here's the kinematics we will draw free body diagram of the system.

09:07 This is c point.

09:08 Here it will be normal reaction friction weight mc into g, mb into g, iv into alpha, ic into alpha, mass of rod into tangential mass of cylinder into tangential and a cylinder into tangential acceleration down finding moment of couple about c point taking clockwise to be positive mcg into r cyanok theta mvg into r cyanox theta mvg into r plus h sine of theta it must be equal to i c into alpha i c into a g tangential into r plus i c into alpha mc into a g into r i v into alpha mc into a g into alpha i v into a p tangential into r plus h from here you make a substitute the value i c is i c into alpha a g tangential is given r into alpha into r i v and the api tangential r plus h alpha r plus h alpha alpha you can take common solving your equation for value of alpha it will be mass of cylinder g into r mass of bar into g into r plus h sin of theta divided by i c mc into r square i b m b into r plus h substituting the value all the parameters are given mass of cylinder is 10 just a moment already we have converted into mass so directly we can substitute mass of point 31056 32 .2 into r 16 by 12 mass of the rod 1 2 232 .32 .2 into r plus h 16 plus 8 upon 16 sign of 20 divided by moment of velocity of the cylinder 0 .27605 plus mass of cylinder 0 .31056 into radius square 16 by 12 whole square plus moment of inertia of the bar point 1 2 422 mass of the war that i have substituted to the war that i have substituted only moment of inertia of the bar is 02875 mass 0 .12 422 into r plus h that is 16 plus 8 divided by 12 whole square so on solving alpha you will get 5 .3896 radiant per second square tangential acceleration of p .24 by 12 into 5 .3896...

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