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(20%/) Problem 2: At its lowest setting centrifuge rotates with an angular speed of 0] 250 rad' When it is switched to the next higher setting it takes t = 11....

Question

(20%/) Problem 2: At its lowest setting centrifuge rotates with an angular speed of 0] 250 rad' When it is switched to the next higher setting it takes t = 11.5 to uniformly accelerate to its final angular speed 02 750 rad 850% Part (a) Calculate the angular acceleration of the centrifuge &1 in rad s` over the time interval

(20%/) Problem 2: At its lowest setting centrifuge rotates with an angular speed of 0] 250 rad' When it is switched to the next higher setting it takes t = 11.5 to uniformly accelerate to its final angular speed 02 750 rad 8 50% Part (a) Calculate the angular acceleration of the centrifuge &1 in rad s` over the time interval



Answers

Centrifuge A centrifuge with a $0.40-\mathrm{kg} \cdot \mathrm{m}^{2}$ rotational inertia has a rotational acceleration of 100 $\mathrm{rad} / \mathrm{s}^{2}$ when the
power is turned on. (a) Determine the minimum torque
that the motor supplies. (b) What time interval is needed
for the centrifuge's rotational velocity to increase from zero
to 5000 $\mathrm{rad} / \mathrm{s} ?$

Well from Newton's second Law for rotational motion, Torque is equal to high times Alfa here eyes, moment off inertia and Alfa is the rotational X relation. So torque is equal to 0.4 multiplied by 100 which is equal to 40 mutant meter right? Well. Given the initial and final velocities and the rotational X relation, we can calculate the time using the laws of motion under constant, angular acceleration. Using omega is equal to omega, not less. Alfa Times T now rearranging for time so Time T is equal to omega minus omega, not divided by Alfa. So omega is 5000 minus zero, divided by 100 so time is equal to 50 seconds.

Part A. We know from the definition of the angular acceleration, this would be the derivative of the angular velocity with respect to time. And so we can then say that the integral from zero to mega times D omega would be equal in the integral from zero to t of alfa d T. And that is how we would find our angular velocity so we can then solve. This would be equaling the integral from zero to t of 5.0 t squared minus 8.5 t d t. And so our angular velocity is then going to be equaling 1/3 times 5.0 by the 5.0, only two significant figures T's t cubed and then minus 1/2 times 8.5 t squared. So this would be your answer for part A for port be. We're doing the exact same thing. Except now we have that we're going to use the interval from zero to say tha d theta uh, well, equal the interval from zero to t of omega d T. Given that the, uh given that the anger, the angular velocity, by definition is the derivative of the angular displacement with respect the time. And so we can then say that this is going to be equal to the square, to the integral of zero to t of. And then we plug what we found in part a So 1/3 times 5.0 t cubed minus 1/2 times 8.5 t squared dt, and we find that Fada is then equaling won over 12 multiplied by 5.0 times T to the fourth power minus one over six times, 8.5 times tea to the third power. This would be your answer for part B and then for part C, they want us to evaluate. At T equals 2.0 seconds. So we can say that Omega at 2.0 seconds with the equaling 1/3 times, 5.0 times and 2.0 cubed minus 1/2 times 8.5 times 2.0 quantity squared. This is equal in approximately negative 3.7 radiance for a second. Fourth, ADA at 2.0 seconds, this would be one over 12 times 5.0 times 2.0 to the fourth power minus one over six times, 8.5 times 2.0 to the third power. And this is equally negative 4.67 radiance. So this would be our answer for the angular velocity. And this would be our answer for the angular displacement. Again, AT T equals 2.0 seconds. That is the end of the solution. Thank you. For what?

So here we're going to use Equation 10 15 forecourt, eh? And we can say that we have 60.0 radiance which would equal 1/2 times omega sub one plus omega sub too multiplied by the time of 6.0 seconds. So, essentially, the angular displacement is gonna be equal to the average angular velocity multiplied by the time, uh, knowing that Omega sub two equals 15.0 radiance per second we confined Omega sub one is equaling 5.0 radiance per second. This would be our answer for part a four part B. We're going to use Equation 10 12. So Equation 10 12 gives that the angular acceleration would then be equal to 15.0 radiance per second, minus 5.0 radiance per second, divided by the time of 6.0 seconds. This is giving us one 0.67 radiance per second squared. This would be our answer for part B four parts. See, we're going to interpret omega equaling Omega Sub one and Beta simply equal state of someone which equals 10.0 radiance. And we can say that the initial angular velocity would be zero radiance for second. Therefore, we can use Equation 10 14 and say that stada initial would be equal to negative Omega sub one square to hide about two times Alfa plus fade us of one and we can then solve and we find that the initial angular displacement is 2.50 radiance. This would be our answer for part C. That is the end of the solution. Thank you for watching.

So the question tells us that a wheel with a diameter of one meter accelerates at a rate of four radiance per second squared. And the first part of the problem wants us to figure out what the final angular velocity is of the wheel at a time T, which is 10 seconds later when were given that the initial angular velocity is to radiance uh, per second. And so to do this, we can use our Kinnah Matic formulas. The one that we're gonna use for a part A is the one that states that the final angular velocity is equal to the initial angular velocity. Plus the acceleration are the angular acceleration times the time. So we know that the initial angular sorry here is the final angular velocity which we're trying to find. We know that the initial angular velocity is to radiance per second plus the, uh, acceleration, which is four times the time, which is 10. So we find that the final, um, angular velocity is equal to 42 radiance per second. So this is the answer to part A. The second part of the problem wants us to figure out what the total angular displacement is equal to. And so to do this, we can just use another kid. A magic formula, which states that the total angular displacement is equal to the initial in your lost e times. The time plus 1/2 times the acceleration times time squares. So we're trying to find this. Tilt the theater here. We know that the initial angular velocity is to times the time, which is 10 plus 1/2 times the acceleration, which is four times the time squared, which is tense, weird. And so if you plug this into a calculator or do you just you could probably just do it in your hand. You'll see that Delta Theta is equal to 220 radiance. So that's the answer to Part B. And lastly, Part C wants us to figure out with the after 10 seconds. What a point on the edge of the ah, the wheel. What it's tangential speed is as well as what the acceleration is of the wheel. So to find a tangential speed, we can take our final, angular velocity, multiply it by the radius of the wheel, and this will give us a tangential velocity so the radius of the wheel is 0.5 meters, right, because the diameter is one meter, so it will just be 42 Radiance per second. Times 0.5 meters gives us 21 meters per second as the tangential velocity. Now, if we want to find the acceleration, there are two places where there's acceleration. There's a tangential acceleration as well as a centripetal acceleration. So to find the tangential acceleration, we can just do the same thing that we did appear for the velocity except will have the angular acceleration times. Uh, the radius is equal to the, um, tangential acceleration. So are angular. Acceleration is for radiance per second squared times the radius which is 0.5 meters and then I give us the conventional acceleration is two meters per second squared now for trying to find what the centripetal acceleration is remembered. This is defined as the, uh tangential velocity squared, divided by the radius squared. So this will be 21 meters per second squared, divided by point five, um, meters. And when you play this to a calculator will get 880 meters per second squared. So now that we have the 10 dental acceleration and the centripetal acceleration, which are perpendicular to each other. We can find the magnitude of the acceleration vector, which is equivalent to the square root of two squared plus 880 squared. And once again, playing this into a calculator gives us a. The magnitude of the acceleration vector is equal to 800 and 82.2 eaters per second squared, and these are your final solutions.


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