a horizontal 90kg merry go round is a solid disc of radius 1.50m and is started from rest by a constant horizontal force of 50N applied tangentially to the edge of the disc the KE of the disc after 3 sec is
Asked by vvelesha
5th September 2018,
10:22 PM
Answered by Expert
Answer:
"Newton's law" for rotational motion, τ = I×α ...............(1)
where τ is torque , which is given by = F×R,
where F is tangential force and R is perpendicular distance between line of action of force and axis of rotation.
I is moment of inertia and α is angular acceleration.
we have, τ = F × R = 50×1.5 = 75 N-m, I = M×(R2/2) = 90×1.5×1.5/2 = 101.25 kg-m2
hence α = 75/101.25 = 0.741 rad/s
angular velocity ω after 3 seconds, " ω = ωo + α×t " ,
where ωo is intial angular velocity that is zero in this question, because the disc starts rotating from rest at t=0.
hence we have ω = 0.741×3 = 2.223 rad/s
Kinetic energy = (1/2)×I×ω2 = (1/2)×101.25×2.223×2.223 = 250 J
Answered by Expert
7th September 2018,
12:31 PM
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