Q. 124.3( 13 Votes )

# 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?

Answer :

We need to find Kinetic energy of the rotating cylinder

We know that, Rotational kinetic energy,

KE = 1/2I ω^{2}

Where

I is moment of inertia

ω is angular velocity

Given

Mass of cylinder, M = 20 kg

Radius of cylinder, R = .25 m = 1/4 m

Angular speed of cylinder, ω = 100 rad s^{-1}

Moment of inertia of a solid cylinder, I = 1/2 MR^{2}

I.e. I = 1/2 × 20 × (1/4)^{2}

= 5/8 kg m^{2}

∴ KE = 1/2 I ω^{2}

= 5/16 (100)^{2}

= 3125 J

= 3.125 kJ

We need to find angular momentum of rotating cylinder about its axis.

Angular momentum, L = I ω

We know that,

Moment of inertia of cylinder, I = 5/8 kg m^{2}

Angular speed of cylinder, ω =100 rad s^{-1}

∴ Angular momentum, L = 5/8 × 100

= 62.5 kg m s^{-1}

= 62.5 J s

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PREVIOUSTorques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.NEXTA. A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.B. Show that the child’s new kinetic energy of rotation is more than the initial for this increase in kinetic energy?

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