The combination of an applied force and a friction force...

The combination of an applied force and a friction force produces a constant total torque of 36.0 $\mathrm{N} \cdot \mathrm{m}$ on a wheel rotating about a fixed axis. The applied force acts for 6.00 s. During this time, the angular speed of the wheel increases from 0 to 10.0 $\mathrm{rad} / \mathrm{s}$ . The applied force is then removed, and the wheel comes to rest in 60.0 s. Find (a) the moment of inertia of the wheel, (b) the magnitude of the frictional torque, and (c) the total number of revolutions of the wheel.

The combination of an applied force and a friction force produces a constant total torque of 36.0 $\mathrm{N} \cdot \mathrm{m}$ on a wheel rotating about a fixed axis. The applied force acts for
6.00 s. During this time, the angular speed of the wheel increases from 0 to 10.0 $\mathrm{rad} / \mathrm{s}$ . The applied force is then removed, and the wheel comes to rest in 60.0 s. Find (a) the moment of inertia of the wheel, (b) the magnitude of the frictional torque, and (c) the total number of revolutions of the wheel.

Key Concepts

Frictional Torque

Frictional torque is the resistive torque that opposes the applied torque in a rotating system. It arises due to friction at the pivot or between moving parts, and it slows down the rotation of the object. In many problems, determining the magnitude of frictional torque is key to understanding how external forces and energy dissipation influence the dynamics of a rotating system.

Torque

Torque is the rotational equivalent of force in linear motion. It represents the tendency of a force to rotate an object about an axis. In rotational dynamics, net torque determines the angular acceleration of an object according to the relation ? = I?, where ? is the net torque, I is the moment of inertia, and ? is the angular acceleration.

Moment of Inertia

The moment of inertia quantifies an object's resistance to changes in its rotational state, similar to how mass resists changes in linear motion. It depends on both the mass of the object and how that mass is distributed relative to the rotational axis. In problems involving rotational acceleration, knowing the moment of inertia is essential for calculating how the applied torques affect the object's angular speed.

Angular Acceleration

Angular acceleration is the rate of change of angular velocity over time. It is determined by the net applied torque and the object's moment of inertia, following the relationship ? = ?/I. This concept is critical in rotational dynamics as it describes how quickly a rotating system speeds up or slows down under a given torque.

Rotational Kinematics

Rotational kinematics deals with the motion of objects in rotation, describing relationships between angular displacement, angular velocity, angular acceleration, and time. These kinematic equations, which are analogous to those in linear motion, enable the calculation of quantities such as total angular displacement (measured in radians or revolutions) and provide insight into the behavior of a rotating system under constant or varying angular acceleration.

Transcript

00:01 In this exercise, we have an object that's rotating due to a torque that is due to a combination of the frictional force and an applied external force.

00:11 The total torque is 36 newton's meters, and it acts on the object for six seconds, taking the object from rest to an angular speed of 10 radiance per second.

00:26 Then the applied force is removed and the object takes 60 seconds to go from 10 radiance per second to rest.

00:39 Then in question a, we have to find the moment of inertia i of the object.

00:48 So remember that the torque, tau, is equal to the moment of inertia i times the angular acceleration.

00:56 And we have the torque and we can calculate the...

00:59 Angular acceleration.

01:02 The angular acceleration alpha can be calculated as the variation in the angular speed divided by the variation in time.

01:14 The variation in angular speed is 10 seconds while, i'm sorry, 10 radiance per second, while the variation in time is 6 seconds.

01:26 So this is 5 third, five thirds radiance per second squared.

01:36 So this means that the moment of inertia will be the torque divided by alpha.

01:42 The torque we were given is equal to 36 newton's meters, and we have to divide it by five thirds radiance per second squared.

01:59 And we obtain 21 .6 kilograms meters squared.

02:14 This is the moment of inertia.

02:18 Then in question b, we have to find the frictional torque.

02:28 So notice that when the applied force is removed, we can calculate the new acceleration, which is equal to my minus 10 radiance per second because the object goes from 10 radiance per second to rest divided by 60 seconds.

02:57 So this is minus 1 6th radiance per second squared.

03:07 And the torque will be just the moment of inertia i times alpha.

03:16 Now i is 21 .6 kilograms meter squared and alpha is minus one sixth radiance per second squared...

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