A heavy flywheel is accelerated (rotationally) by a motor thatprovides constant torque and therefore a constant angularaccelerat
ion Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta. The flywheel is assumed to be at rest at timet = 0 in Parts A and B of this problem.a) Find the time t_1 it takes to accelerate the flywheel to omega_1 if the angular acceleration is Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta.Express your answer in terms ofomega_1 and Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta.b) Find the angle theta_1 through which the flywheel will have turned during thetime it takes for it to accelerate from rest up to angular velocityomega_1.Express your answer in terms of someor all of the following: omega_1, Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta, and t_1.Find the angle theta_1 through which the flywheel will have turned during thetime it takes for it to accelerate from rest up to angular velocityomega_1.Express your answer in terms of someor all of the following: omega_1, Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta, and t_1.c) Assume that the motor has accelerated thewheel up to an angular velocity omega_1 with angular acceleration Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta in time t_1. At this point, the motor is turned off and a brake isapplied that decelerates the wheel with a constant angularacceleration of -5. Find t_2, the time it will take the wheel to stop after the brakeis applied (that is, the time for the wheel to reach zero angularvelocity).Express your answer in terms of someor all of the following: omega_1, Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta, and t_1.
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Answer:
5.24 m/s
Explanation:
So for the rod to be able to rise upward to the straight up position, the kinetic energy caused by linear speed v0 must be just enough to convert into the potential energy.
Since the rod is uniform in mass, we can treat the body as 1 point, at its center of mass, or geometric center, aka 0.35 / 2 = 0.175 m from the pivot.
For the rod to swing from bottom to top, the center must have moved a distance of h = 0.175 * 2 = 0.35 m, vertically speaking.
Since we neglect friction and air resistance, according to the law of energy conservation then:
Where v is the speed at the center of mass, g = 9.81 m/s2 is the gravitational acceleration, and m is the mass. We can divide both sides by m
As this is only the speed at the center of mass, the speed at the bottom end would be different, to calculate this, we need to find the common angular speed:
Where r is the rotation radius, or the distance from pivot point to the center of mass
Where R is the distance from the pivot to the bottom end of the rod