An experimental bicycle wheel is placed on a test stand so...

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axe. If a constant net torque of 5.00 $\mathrm{N} \cdot \mathrm{m}$ is applied to the tire for 2.00 $\mathrm{s}$ , the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 125 s. Compute (a) the moment of inertia of the wheel about the rotation axis; $(b)$ the friction torque; (c) the total number of revolutions made by the wheel in the 125 -s time interval.

Key Concepts

Torque

Torque is the rotational equivalent of force that causes angular acceleration in a rotating system. It is defined as the product of the force and the lever arm distance, and in problems involving constant net torque, it directly relates to the rate at which an object’s rotational speed changes through the equation ? = I?.

Moment of Inertia

The moment of inertia quantifies an object's resistance to changes in its rotational motion about a given axis. It depends on the mass distribution relative to the axis of rotation and plays a crucial role in determining how a torque applied to an object translates into angular acceleration.

Angular Kinematics

Angular kinematics involves the study of rotational motion variables such as angular displacement, angular velocity, and angular acceleration. It uses equations analogous to linear kinematics to relate these quantities over time, allowing the calculation of aspects such as the total number of revolutions completed during periods of acceleration or deceleration.

Friction Torque

Friction torque is the resistive torque produced by frictional forces in rotating systems, such as in the bearings of a wheel. It acts in the opposite direction to the motion and causes deceleration, and is a key factor in determining how long it takes a rotating object to come to rest after an external torque is removed.

Transcript

00:01 Okay, so in this question, we have a bike wheel that is allowed to rotate freely.

00:06 It has a net torque exerted on it of 5 .00 newton meters, and that torque is exerted over two seconds.

00:17 And i'm going to call this time period t1, which is 2 .0 seconds.

00:24 It starts at rest, so its initial omega is 0 .0 .2 .2 seconds.

00:27 It starts at rest, so its initial omega is 0 .0.

00:30 And then its final omega, after this two -second period of time, where it's under this 5 -newton meter torque, it speeds up to 100 revolutions per minute, which if you convert to radiance per second, that is 10 .5 radiance per second.

00:59 You just do that by multiplying the 100 revolution per minute by 2 pi over 60.

01:08 And then after it gets up to this speed, it is allowed to slow down to rest.

01:14 And it does this over a time period, which i'm going to call t2 of 125 seconds.

01:23 So in part a, we're supposed to figure out what the moment of inertia of this bike wheel is.

01:33 And we do this, using the formula, t is equal to i alpha.

01:39 So we are given the torque, but we need to calculate alpha.

01:44 And we can first rearrange this equation to say that i is equal to t divided by alpha.

01:52 And so, yes, we need to calculate alpha because we don't know what that is.

01:56 Luckily, alpha is fairly easy to find.

01:58 It is just equal to the change in omega over t.

02:03 In this case, t1, because we know the change in omega and we know this time period and we know the torque user over this time period.

02:13 So the change in omega, of course, is 10 .5 radians per second and then the time period is two seconds.

02:22 And if you calculate that, you should get that the acceleration alpha is 5 .24 radians per second squared.

02:33 And so we can plug that into this formula here.

02:36 And so the net torque is 5 newton meters over 5 .24 radiance per second squared.

02:46 And if you calculate that, you should get that the moment of inertia of the bike wheel is 0 .955 kilograms meters squared.

02:57 And so that's our answer for part a.

03:00 Now, in part b, we want to find what the torque due to friction is.

03:07 And so now that we know the moment of inertia for the bike wheel, and we can calculate the acceleration again, we can use that to figure out what the torque due to friction is.

03:19 So in part b, we're using the same formula, but we're just using different conditions now...

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