a solid cylinder of mass 1.5 kg and radius 0.1 m rolls down an inclined plan from a height of 3m calculate the rotation energy at the bottom of the plane
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a solid cylinder of mass 1.5 kg and radius 0.1 m rolls down an inclined plan from a height of 3m calculate the rotation energy at the bottom of the plane
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Explanation:
Here,
M = 2 kg
R = 0.1 m
h = 4 m
g = 10 ms-2
Rotational kinetic energy of a body moving with angular velocity ω is given by ½Iω2
Where I = moment of inertia of the body in rotation
For solid cylinder about its central axis,
I = ½MR2
I = ½×2×(0.1)2
= 0.01 kg m2s-2
PE of the cylinder at height h = Mgh
When it reaches the end of the height the PE is entirely converted in to KE. So
½Mv2 = Mgh
=> v2 = 2gh = 2×10×4
=> v = 9 ms-1 (approx.)
Now angular velocity at the foot of the plane is given by
ω = v/R
= 9/0.1
= 90 rad/s
Now
KErot = ½Iω2
= ½ × 0.01× (90)2
= 40.5 J is the Answer.
hope it helps!!