At t=0 a grinding wheel has an angular velocity of 24.0 rad / s It has a constant angular accele

At t=0 a grinding wheel has an angular velocity of 24.0 rad...

$\mathrm{At} t=0$ a grinding wheel has an angular velocity of $24.0 \mathrm{rad} / \mathrm{s}$ It has a constant angular acceleration of $30.0 \mathrm{rad} / \mathrm{s}^{2}$ until a circuit breaker trips at $t=2.00 \mathrm{~s}$. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between $t=0$ and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

$\mathrm{At} t=0$ a grinding wheel has an angular velocity of $24.0 \mathrm{rad} / \mathrm{s}$ It has a constant angular acceleration of $30.0 \mathrm{rad} / \mathrm{s}^{2}$ until a circuit breaker trips at $t=2.00 \mathrm{~s}$. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between $t=0$ and the time it stopped?
(b) At what time did it stop? (c) What was its acceleration as it slowed down?

Key Concepts

Rotational Kinematics

This concept involves the study of rotational motion using equations analogous to those for linear motion. Key equations include theta = theta0 + omega0*t + (1/2)*alpha*t^2 and omega^2 = omega0^2 + 2*alpha*Delta theta, which relate angular displacement, angular velocity, and angular acceleration under constant angular acceleration.

Piecewise Analysis of Motion

When a system undergoes different phases of motion—such as an initial acceleration phase followed by a deceleration phase—it is important to analyze each segment separately. This allows for accurate application of the kinematic equations across intervals with different accelerations before combining the results for overall displacement and time.

Total Angular Displacement

Calculating the total angular displacement requires summing the contributions from each phase of the motion. In problems where a rotating object experiences an initial phase of acceleration and then another phase of deceleration, the overall displacement is the sum of the displacements during both phases.

Deceleration to Rest

This key concept focuses on determining the negative angular acceleration required to bring a rotating system to a halt. It involves setting the final angular velocity to zero and using kinematic relationships to resolve the deceleration and the time taken to come to rest.

Transcript

00:01 So we must find the angular displacement in the first two seconds, how long it took for this entire thing to stop, and then the total time.

00:09 So it would be, we would first write down on givens, the initial angular velocity would be 24 .0 radiance per second.

00:22 And then we have alpha equals 30 radians per second squared.

00:29 And this would occur from two seconds, because that two.

00:34 Seconds the wire trips and actually so we can say that this is at t equals zero and then they want us to find the angular acceleration from two seconds to when to the end of the two in the wheel stops essentially and we also know that delta omega is 532 radians but this is only after t equals two seconds so we should say, well, we have constant acceleration throughout so we can use our kinematic equations applied to rotation, our rotational kinematic equations essentially.

01:25 So this would be theta f equals theta initial plus omega initial t plus half t squared.

01:38 And then we can say omega f, we can actually take this to be zero.

01:43 And then we're going to try, this is going to be for the first two seconds.

01:46 So we can say omega initial is 24.

01:50 This t will be 2 and then plus 1 over 2 .30, 2 squared.

01:58 And this is going to be 108 radians.

02:02 So this would be first two seconds.

02:11 And then if you wanted to find total, it would simply be 108 plus 432, so 540.

02:22 And then we need to find how long it takes.

02:26 So let's get a new workbook.

02:30 We're start off by using this equation.

02:34 We don't know how long it takes, but we should try to find the angular acceleration of the angry acceleration after two seconds.

02:43 So we can say 2 omega delta theta.

02:52 At this point, let's solve for angular acceleration.

02:55 So angular acceleration would equal minus omega -mega.

02:59 All over.

03:04 Here our final is going to be zero because we are stopping and then our initial will simply be equal to the final velocity after two seconds...

Please give Ace some feedback