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A cylinder with a moment of inertia, I = 4 kgm2 is rotating with an angular velocity of 6 rad/secwhen a brake applies a torque to slow the cylinder to a stop inonl...

Question

A cylinder with a moment of inertia, I = 4 kgm2 is rotating with an angular velocity of 6 rad/secwhen a brake applies a torque to slow the cylinder to a stop inonly 12 seconds. (a) What is the angular acceleration of thecylinder? (b) What is the torque provided by thebrake?Please help Thank you!

A cylinder with a moment of inertia, I = 4 kg m2 is rotating with an angular velocity of 6 rad/sec when a brake applies a torque to slow the cylinder to a stop in only 12 seconds. (a) What is the angular acceleration of the cylinder? (b) What is the torque provided by the brake? Please help Thank you!



Answers

A $5.0 \mathrm{kg}, 60$ -cm-diameter disk rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. a. What is the cylinder's initial angular acceleration? b. What is the cylinder's angular velocity when it is directly below the axle?

Yeah. Here for the solution for the party. The vertical displacement before coming to rest the six metre calculate the angular displacement the drum will have in the process of coming to rest. Using the equation that heat equal to s by are here, as is the distance traveled by the cylinder and R is the radius of that room. Now substitute 6 m for s and 2 15 m for our and the question. So we get to take or to six metre by 2. 50 multiplied by 1 m by 1000. Mm. So from here we get angular velocity to take all to 24 ingredients. Now calculate the angular velocity of the drum. Initially using the equation Omega not equal to we buy are here vis the linear velocity of the drum. Initially. Now substitute 3 m per second for V and 2 50. 4 are here. By substituting the value, we get omega nautical to 3 m per second. Divided by 2. 50. Multiply by 1 m by 1000. Mm. And from here we get omega not equal to 12. Red purse, Second radiance per second. Since the drum comes to rest. The final, angular velocity of Trump is zero now Calgary The angular acceleration of the drum using the equation, Omega Square minus Omega Nordisk are equal to two alpha Tita. Here, Alpha is the angular acceleration, and omega is the final angular velocity. Now substitute 24 radiance, 40 to 12 radian per second for omega Note and zero for omega here. By substituting the value, we get zero square minus 12 historical to help for multiplied by 24. From here we get angle acceleration. It is difficult to minus three radian per second square. The angular acceleration is negative, implies the drum is decorating. Therefore, the angular acceleration of the trump is three radian per second square. Now for the part B. Calculate the time required for the drum to come to rest, using the equation that omega equal to omega, not plus 80. Now substitute minus three rate per second. Squared for alpha 12 8 per second for a big A note and zero for omega in the equation. Omega quartz, Omega net plus alpha T. So have I, substituting the value we get zero equal to 12 minus three multiplied military and from her media time equal to four seconds. Therefore the time required for the drum to come to the rest is for second. Now, this is a complete solution with the concept and the explanation. Step by. Step in detail. Please go through this. Thank you.

So we need to calculate the torque by the motor. The mass of the grindstone is one point six kilograms. We're going to model the grindstone as a uniform cylinder with the radius of point two zero meters are made. A final is twenty four revolutions per second. And we can say that our mega initial is a zero radiance per second. We're going immediately convert. And so the omega final equals forty eight pi radiance per second. And we can say that Delta T equals six point zero seconds. So at this point, we can find that we know that the moment of inertia of a cylinder she's getting equal r squared over two. So at this point, we can say that we can say that town is going to be equal to ay Alfa and Alfa. In this case will be Delta Omega over Delta T and initially it zero. And finally, it's twenty four angular velocity final twenty four revolutions per second or forty eight pi ratings for a second. So we can just say that this is going to be equal to Omega final over now to the teeth and Tao or torque is going to be equal to, um r squared Omega Final over two. Thirty. And we can solve and say that Tao is going to equal one point six point two zero squared forty eight pi divided by two times six point. Oh, and this is going to be point eight zero for Newton meters. So this is the, um, required torque of the rotor so that the grindstone can reach twenty four revolutions per second in the time interval of six seconds. So that would be your answer. And that's the end of the solution. Thank you for watching.

In this problem we have to determine the complete revolutions. Uh does the cylinder retake? So the expression for angular excavation is 81. Need a phone too. That is twice 80 upon the and the expression for angular displacement. That is theta is Omega -0mega notes square upon to ice. The angular acceleration. So omega noticed that initial angular velocity and terrorist angular displacement. I don't know. Let us put the value here and make us square minus omega not square. I want to wife to, to to multiplied by two weeks for a day. A point The So given we have omega is 1.2. Waiting for a 2nd. My evidence to the point 05 radiant for a second. Square Upon four. multiplied by uh zero point. There's no one made meter for a second. Square upon 20 santa Monica. So that is a call to 36.5 radiant. I converted to Invitation. That is one rotation and twice by video. There is a cult of 5.8 rotations. So therefore number of complete revolution. Because the cylinder the date. Yes. Five flirtation

This problem covers the concept of the uh better access to them and the conservation of energy. So to solve this problem for part a, we need to write the equation by using the parallax history. Um so the rotational inertia about tax interpretation of the cylinder equals the rotational and I say about a central Alexis that is mm R squared up on two plus the mass of the cylinder into the square of the distance of the center of mass from the X interpretation that is being square. I also understood the value. So The mass of the surrender is 20 KK into The radius is 0.1 m square upon to plus 20 kg into the distance from the exploitation. This five centimeter, We can write five into 10 days -2 m square of the additional inertia of the cylinder. About the successes 0.15 Kg metal square no party. So in this part we cover the concept of the conservation of energy, which means the initial energy is the Cuban. It'll the final image. And when the cylinder is at the bottom position, that is at this position, the changing the position of center of mass, The vertical position is to D. So initially the surrender has a um potential energy. So we can bait the mass of the surrender m energy into H. And that equals the final kind of technology. That is um half of the rotational inertia into, oh my God, square. And when we know that H equals two times the the intensity. So we can write MG into to the equals half of I. Omega is quick. So the angle speed at the bottom of the uh at the bottom point is squared off four MG. D. Upon the rotation in us. Here, I now let's substitute the value. So omega Equals Square root of four times 20 kg and to 9.8 m for 2nd square into 0.05 meter upon 0.15 Kg meter square, the angry speed at the bottom is 16 radiance or second.


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