A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular accelerationalpha. The flywheel is assumed to be stationary at time t = 0 in Parts A, B, and C of this problem.

a). Find the time t_10 it takes to accelerate the flywheel to 10.0 rps (revolutions per second) if alpha is 5.00 radians/s^2.

b). Find the time t to accelerate the flywheel from rest up to angular velocity omega_1.
Express your answer in terms of alpha and omega_1.

c). Find the angle theta_1 through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity omega_1.
Express your answer in terms of any of the following: omega_1, alpha, and/or t_1.

d). Assume that the motor has accelerated the wheel up to an angular velocity omega_1 with angular acceleration alpha in time t_1. At this point, the motor is turned off, and a brake is applied that decelerates the wheel with a constant angular acceleration of -5alpha. Find t_2, the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity).
Express your answer in terms of any of the following: omega_1, alpha, and/or t_1.

Solution Summary

The solution analyzes the angular motion of the flywheel, such as the angular acceleration, angular velocity, and time.

... The equation of motion for the angular velocity is: ω = ω 0 − αt. Or: ω − ω0 t=− α Since initially the flywheel rotates with angular speed ω0, the ...

Rotational Motion: Flywheel. ... To determine the MI of the flywheel we integrate (1) with limits r = 0.9R and r = R ...Angular speed ω = 2Пf = 2П x 5 = 31.4 rad/sec. ...

... is the kinetic energy of the flywheel after charging ? ... In rotational motion, the angle plays the role of ... time transform into angle = angular velocity × time. ...

... Final angular velocity of both the flywheels is given as. ... For flywheel (B ... a) Total kinetic energy of the system = Energy due to translational motion of container ...

... 0 c) From Newton's second law applied to rotational motion : Torque = Rate of change of angular momentum. ... of weight mg, the axis of the flywheel will tend ...

1. At t=0, a fly wheel has an angular velocity of 4.7 rad ... 1. At t = 0, a flywheel has an angular velocity of ... Solution The basic equations for angular motion are. ...

... A mass of 0.5 kg is suspended from a flywheel as shown in FIGURE 2 (attached). ... (b) The angular acceleration of the ... (d) The frictional torque, resisting motion. ...

... velocity and instead of linear acceleration a we use angular acceleration . Therefore the equations of motion are: An engine flywheel initially spinning ...

KE/rotational motion. ... So the kinetic energy stored in the flywheel is. ... V is the linear velocity, which we have to find out from the angular velocity of fly wheel. ...