Asked by Gift
A constant force of 40newton is applied tangentally to the rim of the wheel with 20cetimetre radius the wheel hs moment of intial 30kg find (a)angular acceleration (b)angular speed after 4second frm rest (c)d number of revolution made in(d)show that d work done of d wheel in dis 4seconds is equal to d kinetic energy of d wheel after 4seconds.
Answers
Answered by
Damon
Seems to be a bad typing day
hs moment of intial 30kg
Suspect you mean moment of inertia but the units are wrong.
If it is I = 30 kg m^2
then torque = I alpha
alpha = angular acceleration
= 40 * 0.20 N m / 30 kg m^2
omega = alpha * t
theta = (1/2) alpha t^2
number of revs = theta/2pi
work done = Torque * theta
should be (1/2) I omega^2
hs moment of intial 30kg
Suspect you mean moment of inertia but the units are wrong.
If it is I = 30 kg m^2
then torque = I alpha
alpha = angular acceleration
= 40 * 0.20 N m / 30 kg m^2
omega = alpha * t
theta = (1/2) alpha t^2
number of revs = theta/2pi
work done = Torque * theta
should be (1/2) I omega^2
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