A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Mass of the cylinder, m= 20 kg
Angular speed, ω = 100 rad s-1
Radius of the cylinder, r= 0.25 m
The moment of inertia of the solid cylinder:
I = mr2 / 2
= 1/2 x 20 x (0.25)2
= 0.625 kgm2
∴Kinetic energy = 1/2 I ω2
= 1/2 x 6.25 x 1002 = 3125 J
∴Angular momentum, L= Iω = 6.25 × 100 = 62.5 Js
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(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
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(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
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(b) What is the force of friction after perfect rolling begins?
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(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
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