A disk rotating in a vertical plane about a horizontal axis through its center,

Question

A disk rotating in a vertical plane about a horizontal axis through its center, has a cord wrapped around its outer edge and connected a mass m1=0.200kg. The disk has a mass M=0.800kg and a radius r=0.150m. The system is released from rest (call this position (1)) and the hanging mass falls a distance of 0.600m (call this position 2).

a) Use the work energy methods and calculate the velocity of the hanging mass at position 2.

b) What fraction of the total energy of the system is the rotational energy of the disk when position (2) is reached?

c) Calculate the net work done on the system from position (1) to position (2).

(Answers: a) 1.98m/s, b) 0.667, c) 1.18J)

Explanation / Answer

The gravitational potential energy lost by the mass must be equal to the kinetic energy gained by the mass and the disc. Let the mass of the disc be M= .8 kg and small mass m= .2 kg. The height through which it falls down is h= .6 m .then

mgh = (mv2) /2 + (Iv2) / (2r2) Where (Iv2) / (2r2) is the rotational kinetic energy of disc and   (mv2)/2 is the translational kinetic energy of small mass.

Or. v = [ (2mgh) / (m+(I/r2) )]1/2

On substitution we get velocity of mass at position 2 as v= [ (2x.2x9.8x.6) / (.2+( .8/2))]1/2 = 1.979 m/s

Where I is the moment of inertia of disc which is I = (mr2)/2 where r is the radius of disc

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