A half-cylinder with mass m and radius r is released from rest in the position shown. Knowing th

A half-cylinder with mass m and radius r is released from...

Key Concepts

Rolling Without Sliding

This concept refers to the condition where the point of contact between the rolling object and the surface has zero instantaneous velocity relative to the surface. It establishes a fixed relationship between the angular velocity of the rolling object and its translational velocity, implying that the distance traveled by the object is directly proportional to the angle through which it rotates.

Conservation of Energy

In problems involving rolling objects, the principle of conservation of energy is used to relate the initial potential energy to the final kinetic energy. Here, the gravitational potential energy lost by the object as it descends is converted into both rotational and translational kinetic energy, enabling the computation of quantities like angular velocity after a given movement.

Moment of Inertia and the Parallel-Axis Theorem

The moment of inertia quantifies an object's resistance to angular acceleration about a given axis, and its calculation often involves the parallel-axis theorem. This theorem is essential when the axis of rotation does not pass through the center of mass, allowing one to adjust the known moment of inertia about the center of mass by adding a term that accounts for the distance between the axes.

Reaction Forces

Reaction forces include the normal and frictional forces exerted by the surface on a rolling object. These forces ensure that the object does not sink into or slide across the surface and play a significant role in the dynamics of rolling motion. The normal reaction force is particularly important in assessing balance and equilibrium in such systems.

Center of Mass Considerations

The location of the center of mass in a body affects its potential energy and the dynamics of its rotation. For objects with non-uniform or asymmetric shapes, such as a half-cylinder, the center of mass is not at the geometrical center. Calculating its position is crucial for accurately applying energy conservation and computing the moment of inertia about the correct axis.

Transcript

00:01 Hello friend in this problem it is given a half cylinder having the mass and radius r is released from rest.

00:10 So omega 1 is 0 in the position shown in figure if it rolls without sliding that is for pure rolling motion calculate first part angular velocity if it rolls through 90 degree second reaction.

01:03 Reaction at the horizontal surface at the same extent.

01:09 Reaction we have to calculate at horizontal surface for theta to be 90 degrees.

01:31 Let us start solving it.

01:37 Remembering it, this distance will be go 4 r upon 3 pi and its moment of inertia will be half of m r square minus m into g o square applying parallel x theorem for negative mass system this is position 1 this is position 2 oh this is g now this is o and this is g and this distance we know og now position 1 it is at rest so kind of energy is zero considering the potential energy to be zero position two as shown in the figure potential energy is minus m g into og that is minus m g is 4 r by 3 pi so we can write 4 m g r upon 3 pi now kinetic energy first we will measured already i have written moment of inertia half mr square minus og og is 4 r upon 3 pi into m on solving it you will get point 3198 7mr square kinematic for the point c velocity v is called to b into omega 2 that is value of v will be r minus o g so 4 r upon 35 into omega 2 substitute in the value r is given our value of r is not known so in terms of our this value can be calculated by taking r to be common you will get 0 .57 559 r omega 2 so chyletic energy and position to v2 half m b squared plus half i omega square i is 0 .31 9873 omega square we can substitute the value of v2 half m 0 .57559 or omega 2 square here there is a mistake in substituting the i that is mr square omega square so half point 319873 m r square omega square so on solving it you will get 322 to be 0 .32 559 mr square omega 2 square now applying the conservation of energy initial kinetic energy plus initial potential energy position of 1 plus position of 2.

08:21 T1 is 0, potential energy is 0...

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