A pulley on a frictionless axle has the shape of a uniform...

A pulley on a frictionless axle has the shape of a uniform solid disk of mass $2.50 \mathrm{~kg}$ and radius $20.0 \mathrm{~cm}$. A $1.50 \mathrm{~kg}$ stone is attached to a very light wire that is wrapped around the rim of the pulley (Fig. E9.47), and the system is released from rest. (a) How far must the stone fall so that the pulley has $4.50 \mathrm{~J}$ of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

A pulley on a frictionless axle has the shape of a uniform solid disk of mass $2.50 \mathrm{~kg}$ and radius $20.0 \mathrm{~cm}$. A $1.50 \mathrm{~kg}$ stone is
attached to a very light wire that is wrapped around the rim of the pulley (Fig. E9.47), and the system is released from rest. (a) How far must the stone fall so that the pulley has $4.50 \mathrm{~J}$ of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

Key Concepts

Relationship Between Linear and Angular Quantities

In systems with interconnected translational and rotational motion, like a string unwinding from a pulley, the linear velocity of the falling mass is directly related to the angular velocity of the rotating object (v = ?r). This relationship is critical for connecting the energy transformations of translational and rotational motion.

Moment of Inertia

The moment of inertia represents an object's resistance to changes in its rotational motion and depends on the mass distribution relative to the rotation axis. For a uniform solid disk, a standard form of the moment of inertia is used, which is essential for calculating the rotational kinetic energy in problems involving rotating bodies.

Conservation of Energy

This principle states that the total energy of an isolated system remains constant despite transformations between different forms, such as potential, translational kinetic, and rotational kinetic energy. In problems involving a falling mass attached to a rotating object, the loss in gravitational potential energy is converted into both translational and rotational kinetic energies.

Rotational Kinetic Energy

Rotational kinetic energy is the energy stored in a rotating object, given by the expression ½I?², where I is the moment of inertia and ? is the angular velocity. It is a key concept when analyzing systems where objects rotate, such as pulleys, and helps in understanding how energy is distributed between rotating and non-rotating parts of a system.

Transcript

00:01 For this problem, we're looking at a weight that has been attached to a pulley by a mass of string.

00:10 The pulley is being pulled by gravity, and we are supposed to figure out how far it has to fall to import 4 .5 joules of energy into the pulley.

00:25 So we know right off the bat that the energy going into the system is going to be equal to mgh.

00:32 This m is the mass of the weight.

00:36 Since it is what's being affected by gravity, it's the thing moving while the pulley is remaining stationary.

00:45 And that is going to be equal to the two places where the energy is going.

00:50 Because of the conservation of energy, we know that's going to be equal to the two energies added up.

00:56 The only places it can go are into the velocity of the weight, which is one -half mv squared, and into the rotation of the pulley, which is one -half moment of inertia times the angular velocity squared.

01:12 Now because we know that this pulley is a disk, we can translate these into similar terms.

01:20 1 -5mv squared plus 1 -half times the moment inertia of a disk is 1 -half big m for this one.

01:32 This is the mass of the pulley, r squared, times the angular velocity, which is, of course, equal to v over r, but the angular velocity is squared, so i'll square this two.

01:47 And this little r is the radius of the pulley, so i can do that.

01:52 And then these two will cancel, and what we're left with is that this is equal to 1⁄2 mv squared, plus mv squared over four i may as well continue simplifying this this is going to be equal to v squared over two times m plus big m over two so the next thing we need to do uh we have all of these terms except for h of course which is what we're looking for and v squared uh but we can find squared because we know what the energy we're looking for in the pulley is and we know that the energy in the pulle is can be equal to this term big m v squared over 4 e1 is equal to 4 .5 which is equal to uh no i'll just do that at the end it's equal to mv squared over 4 and now to solve for v we can flip this around pull v out on the left v squared is equal to 4e1 over big m which is equal to 4 times 4 .5 over 2 .5 which will give us a velocity squared of 7 .2 meter squared per second squared and now we can plug this in here.

03:55 I'm going to simplify down one further.

03:58 I'm going to pull out this m .g.

03:59 Divided out so that i'm left with h alone...

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