A constant torque of 500Nm turns a wheel about its center. the moment of inertia about this axitis 100kgm.find the angular velocity gained in 4 second and kinetic energy gained after 20 revolutions
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1. A constant torque of 200N m turns a wheel about its centre. The moment of inertia about this axis is 100kg m^2. Find
(a) the angular velocity gained in 4s
Torque = inertia x alpha (alpha is angular acceleration, rad/s^2)
200 N.m / 100 kg.m^2 = 2 rad/s^2
angular speed = omega (rad/s) = alpha x time = 2 x 4 = 8 rad/s (answer)
(b) the kinetic energy gained after 20 revs
work done by torque = torque x distance in radians.
20 revs x 2pi = 125.66 radians
torque = 200 N.m
200 x 125.66 = 25132.74 J (answer)
2. A flywheel has a kinetic energy of 200J. Calculate the number of revolutions it makes before coming to rest if a constant opposing couple of 5N m is applied to the flywheel.
as above, use work done by torque, but in this case it will be work done by frictional torque, the 5 N.m should actually be -5 N.m, but still, i'll stick to the question.
200/5 = 40 radians
40/(2pi) = 6.366 revolutions (answer)
3. If the moment of inertia of the flywheel about its centre is 4kg m^2, how long does it take to come to rest.
Rotational KE = 200 J
RKE = 1/2Iω^2
200/(0.5 x 4) = ω^2 = 100, sq-root(100) = ω = 10 rad/s
40 radians / ((10 + 0)/2) = time = 8 secs