A small block on a frictionless, horizontal surface has a...

A small block on a frictionless, horizontal surface has a mass of 0.0250 $\mathrm{kg}$ . It is attached to a massless cord passing through a hole in the surface (Fig. 10.48 ). The block is originally revolving at a distance of 0.300 $\mathrm{m}$ from the hole with an angular speed of 1.75 $\mathrm{rad} / \mathrm{s}$ . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 $\mathrm{m}$ . Model the block as a particle. (a) Is angular momentum of the block conserved? Why or why not? (b) What is the new angular speed? (c) Find the change in kinetic energy of the block, (d) How much work was done in pulling the cord?

Key Concepts

Angular Momentum Conservation

In a system where only central forces act, the net external torque about the pivot point is zero, which means that angular momentum remains constant throughout the motion. This concept is important because even if the radius of rotation changes, as long as the forces acting (like the tension in the cord) do not produce a torque about the center, the product of the moment of inertia and the angular speed remains unchanged.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. For a particle moving in a circle, it is proportional to the mass of the particle and the square of its distance from the axis of rotation. When the radius decreases, the moment of inertia decreases as well, which, under angular momentum conservation, leads to an increase in angular speed.

Work-Energy Principle

The work-energy principle states that the work done by external forces on an object is equal to the change in its kinetic energy. In the context of changing the radius of rotation, even though the angular momentum is conserved, work is done to pull the cord, which results in a change in the kinetic energy of the rotating block.

Centripetal Force and Radial Motion

In circular motion, centripetal force is required to keep an object moving along a curved path, acting towards the center of the circle. When the radius of the rotation is altered by pulling the cord, the tension acts radially and provides the necessary centripetal force with no torque about the center, thereby enabling conservation of angular momentum while changing other parameters such as angular speed and kinetic energy.

Transcript

00:01 Okay, so in this question, we have a block tied to a string, and this block is spinning in a circle on a frictionless table.

00:08 And initially, the mass of this block, well, the mass doesn't change, but the mass of this block is 0 .0250 kilograms.

00:18 And the initial radius of the circle that this block is spinning in, our eye, is going to be 0 .330 meters.

00:30 And the initial rotational speed, omega initial, is 1 .75 radians per second.

00:43 And then, you know, it's a physics problem.

00:46 Nothing stays the same forever.

00:47 So the string is pulled.

00:50 The string is, you know, going through a hole at the center of this circle.

00:54 The string is pulled to make the radius of the circle smaller.

00:57 And so the final radius, our f, is 0 .150 meters, half of what we had before.

01:07 And in part a, we're asked, is angular momentum conserved? and why or why not? so the answer here is yes.

01:20 And the reason is there is no external torque applied to this system.

01:26 The only force that's applied to the system is the pulling of the string, which radially goes inward to the center of the circle.

01:36 So there is no, there is a force applied, but the angle between the force and the lever arm is zero.

01:44 So there is no torque applied here.

01:47 And since the net torque applied to a system is equal to the change in angular momentum over time, if the net torque is zero, since there's no torque applied, we know that torque is equal to zero, then the change in angular momentum is zero.

02:07 So since the change in angular momentum is zero, then we know that angular momentum is conserved.

02:14 So the reasoning, the answer is yes.

02:17 The reason is since there's no net torque supplied, then there is no change in momentum.

02:27 So angular momentum is conserved.

02:34 And so now we can move on to part b.

02:38 And here we're asked to find what the final angular speed is after the string has been pulled and the change in radius has happened.

02:48 So to do to do this, we set up the equation the initial angular momentum is equal to the final.

02:53 And then the initial angular momentum has the formula of i initial times omega initial is and then the that's equal to the final angular momentum.

03:03 So i final, final is our sorry, times the final omega.

03:13 And we're told to model this box that's spinning around as a particle.

03:19 So the formula for i is just going to be mr squared.

03:23 So our equation then becomes mr initial squared times omega initial.

03:31 It's equal to mr final squared times.

03:35 Times omega final.

03:37 And of course the mass of the box stays the same...

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