A uniform cylinder of radius 10 cm and mass 20 kg is mounted so as to rotate freely about a ho

A uniform cylinder of radius 10 cm and mass 20 kg is...

A uniform cylinder of radius $10 \mathrm{~cm}$ and mass $20 \mathrm{~kg}$ is mounted so as to rotate freely about a horizontal axis that is parallel to and $5.0 \mathrm{~cm}$ from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

A uniform cylinder of radius $10 \mathrm{~cm}$ and mass $20 \mathrm{~kg}$ is mounted so as to rotate freely about a horizontal axis that is parallel to and $5.0 \mathrm{~cm}$ from the central longitudinal axis of the cylinder.
(a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

Key Concepts

Rotational Inertia

Rotational inertia, also known as moment of inertia, is a measure of an object’s resistance to changes in its rotational motion about a given axis. It depends on both the mass of the object and how that mass is distributed relative to the rotation axis. For standard geometric objects, formulas have been derived to express this relationship, and these formulas serve as starting points for further analysis when the axis of rotation is shifted or altered.

Parallel Axis Theorem

The parallel axis theorem is a useful tool in rotational dynamics that enables the calculation of an object's moment of inertia about any axis, given its moment of inertia about a parallel axis through its center of mass and the perpendicular distance between these axes. This theorem is essential when dealing with systems where the axis of rotation does not pass through the center of mass, as it accounts for additional contributions from the mass distribution relative to the new axis.

Energy Conservation in Rotational Motion

In systems involving rotation, mechanical energy conservation principles extend to include both translational and rotational forms of energy. When an object rotates under the influence of gravity without energy loss, its loss in gravitational potential energy is converted wholly into rotational kinetic energy, which is expressed in terms of its angular speed and moment of inertia. This principle facilitates the calculation of angular speed or other dynamic quantities when the system's energy changes.

Transcript

00:01 I'm working on this problem, and i wrote down the radius and the mass, and the distance, which is right in here, is the distance from the axis of rotation to the center of mass.

00:16 Now, i know that the inertia is going to be the inertia around the center of mass, which is one half mr squared because it's a cylinder, which is like a disk.

00:24 But then i have to add md squared to move the axis, putting that in a calculator.

00:36 It disappeared okay well r equals 0 .1 m equals 20 d equals 0 .05 so i is going to be 1 half m r squared plus m d squared so that gives me the 0 .15 kilogram meter squared now, in part b, i'm going to draw the end view and the green line, the axes of rotation is like this in this description of part 2.

01:25 Nevertheless, the force that acts on this is mass times acceleration due to gravity in its center of mass.

01:33 So the torque is going to be mass times acceleration due to gravity times the distance.

01:41 Which is d, which is going to equal i alpha.

01:49 So alpha is mgd over i.

01:56 By the way, the mass cancels out there.

02:00 Nevertheless, i also know that omega final squared equals omega -initial squared plus 2 alpha delta theta.

02:16 Delta theta, we just want this to rotate downward 90 degrees.

02:24 So delta theta 90 degrees, that would be pi over 2 radiance.

02:36 So omega would be the square root of 2 alpha times pi over 2, which is just going to be the square root of alpha pi...

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