Solve Prob. 16.30 , assuming that the direction of motion of...

Key Concepts

Vector Directions and Sign Conventions

In physics problems involving motion, it is crucial to establish a consistent sign convention. When the direction of motion is reversed, the signs of velocity, acceleration, and forces must also be reversed accordingly. This concept is fundamental in ensuring that equations of motion and force balance are correctly applied.

Newton's Second Law

Newton's Second Law is the foundation for analyzing dynamics in any mechanical system. It relates the net force acting on an object to its mass and acceleration. In problems where the direction of motion is reversed, the application of F = ma requires careful attention to the signs of either the acceleration or the forces involved.

Belt and Pulley System Dynamics

Belt and pulley systems are common in the transmission of power and motion. Analyzing these systems involves understanding how forces, tensions, and friction are distributed along the belt as it moves over pulleys. Reversing the direction of the belt can alter the direction in which these forces act, thereby affecting the overall dynamics and performance of the system.

Friction Forces

Friction plays a key role in belt-driven mechanisms since it resists relative motion between contacting surfaces. The frictional force always acts opposite to the direction of contact motion. Thus, when the belt’s direction is reversed, the frictional force also reverses, which must be taken into account during the analysis of the system’s equilibrium or motion.

Transcript

00:01 Hi everyone here it is given a disk of radius 180 mm.

00:07 This is the radius of the disk which is 180 millimeter that is 0 .18 meter is at rest when it is placed so initially it is at rest when placed in contact with the belt moving downward with speed be neglecting the weight of the link a -v to be negligible coefficient of friction between the belt and the disk is 0 .4 we have to find angular acceleration when silping starts this will be the direction of normal reaction there will be a friction and its weight m g now let us start performing see the free body diagram again friction on the disc will be in this direction on the well it will be in the upper direction normal reaction in this direction weight vertically downward on the disk so let us see here submission of forces along a horizontal direction must be zero because it is in translatory equilibrium so n minus f a v there will be a force i forget to write here the force on a will be in this direction along a v direction i have marked wrongly just a moment this will be the direction of force now and minus a v f a v cause of theta so from here you write f a v cause of theta to be n first equation now summation of force along by axis upward we are taking positive that must be also zero so f a v sign of theta minus m g minus friction and friction will be mu k into n so from here we can write f a b to be f a v sine of theta to be m g plus mu k and divide equation 2 by 1 so you will get 10 of theta to be m g plus mu k and divide equation 2 by 1 so you will get 10 of theta to be to be m g plus mu k n upon n so from here value of n you will get m g upon 10 of theta minus mu k so you may calculate the friction the friction will be mu k n that is mu k m g divided by 10 theta minus mu k so this is equation now due to the friction it will experience the disc experience the angular acceleration so moment of force on that is about the point a you will get friction into r is equal to i into alpha so from here you may write alpha to be r upon i into f r is this moment of the disk about its perpendicular axis mr square and forces mu k m g divided by 10 of theta minus mu k simplifying so you will get alpha to be 2g up on r into mu k upon 10 of theta minus mu k now substituting the value g is 9 .8 radius is 0 .18 coefficient of friction is 0 .4 10 of 60 minus 0 .4...

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