A thin, light string is wrapped around the outer rim of a...

A thin, light string is wrapped around the outer rim of a uniform hollow cylinder of mass 4.75 kg having inner and outer radii as shown in $\textbf{Fig. E10.25.}$ The cylinder is then released from rest. (a) How far must the cylinder fall before its center is moving at 6.66 m/s? (b) If you just dropped this cylinder without any string, how fast would its center be moving when it had fallen the distance in part (a)? (c) Why do you get two different answers when the cylinder falls the same distance in both cases? Figure E10.25 (CAN'T COPY THE FIGURE)

A thin, light string is wrapped around the outer rim of a uniform hollow cylinder of mass 4.75 kg having inner and outer radii as shown in $\textbf{Fig. E10.25.}$ The cylinder is then released from
rest. (a) How far must the cylinder fall before its center is moving at 6.66 m/s? (b) If you just dropped this cylinder without any string, how fast would its center be moving when it had fallen the distance in part (a)? (c) Why do you get two different answers when the cylinder falls the same distance in both cases?
Figure E10.25
(CAN'T COPY THE FIGURE)

Key Concepts

Moment of Inertia

The moment of inertia measures how resistant an object is to changes in its rotational state. It depends on the mass distribution relative to the axis of rotation. For a hollow cylinder, the moment of inertia plays a crucial role in determining how much energy is diverted into rotational motion compared to translational motion as the cylinder falls.

Rolling and Unwinding Constraints

When an object rolls without slipping or unwinds a string, there is a kinematic constraint linking linear speed and angular speed (typically expressed as v = ?r). This relationship is essential in problems where both rotational and translational motions occur, because it allows one to translate between the two forms of motion and correctly apply energy conservation.

Conservation of Energy

This principle states that in an isolated system, the total energy remains constant. When an object falls, its gravitational potential energy is converted into kinetic energy. In problems involving both rotation and translation, the lost potential energy is partitioned into translational kinetic energy and rotational kinetic energy. This concept is used to determine the speeds or distances involved in the motion.

Rotational Kinetic Energy

Rotational kinetic energy is the energy an object possesses due to its rotation. It is determined by the object's moment of inertia and angular velocity. When a cylinder unwinds or rolls, part of the gravitational potential energy goes into spinning the cylinder, making it important to account for this energy when analyzing its overall motion.

Translational Kinetic Energy

Translational kinetic energy is the energy associated with the linear motion of an object. When an object is in free fall, all of its gravitational potential energy is converted into translational kinetic energy. This is contrasted with cases where some energy is also converted into rotational kinetic energy, resulting in different speeds for the same vertical displacement.

Transcript

00:02 So in this problem we have a hollow cylinder.

00:09 Now, um the figure doesn't give us the values like it should, but we can still solve this problem by plugging in variables.

00:17 So are one for the inner radius and are too for the outer radius.

00:21 And we're told that it has a mass of 4.75 kg.

00:30 Yeah.

00:30 And so it asks us, how far do we need to drop it for? the center is moving at 6.66 m for a second.

00:40 So the first thing i see when i think of this conservation of energy, it's the first thing we do is take the potential energy, which is mgh set it equal to the kinetic energy.

00:50 Now there's two components to the kinetic energy here.

00:52 There's a normal one from the linear kinetic energy one half of b squared and there's a rotational one half at the moment of inertia.

01:00 And omega the angular velocity squared.

01:04 Now we know i the moment of inertia for a hollow sphere is one half mm times the inner radius squared plus the outer radius squared.

01:15 Ok.

01:16 And you can derive this or you can just get it out of a book usually will be given to you.

01:21 Okay? and we know omega is equal to linear velocity over radius.

01:27 Okay.

01:27 And in this case, since we have this string on the outer diameter here, the radius here is going to be hard to.

01:36 Okay, so you can plug this in m g h equals one half mv squared plus one half i whoops.

01:51 We're supposed to plug this in uh one half m.

01:57 R one squared plus r.

02:00 Two squared at times.

02:04 Omega square it's a v squared over r two square.

02:08 Okay now we can plug in values for mass.

02:12 So 4.75 and she is 9.8 one.

02:19 And then height is what we're trying to find equals one half 4.75 times.

02:29 And the velocity we're trying to get to is 6.66 plus 1/4.

02:39 4.75...

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