Q. 193.8( 16 Votes )

# A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?

Answer :

What we want to find is work done. We know that the work done to stop a moving body is the total kinetic energy of the body.

∴ Work Done = KE = KE_{rot} + KE_{tran}

Where KE_{rot} is rotational kinetic energy and

KE_{tran} is translational kinetic energy

KE_{rot} = 1/2 Iω^{2}

KE_{tran} = 1/2 Mv^{2}

Where M is mass of hoop

I is moment of inertia of hoop

v is velocity of centre of mass

ω is angular velocity

Now the given values are,

Mass of hoop, M = 100 Kg

Radius of hoop, R = 2 m

Speed of centre of mass of hoop, v = 20 Cm s^{-1}

= .2 m s^{-1}

The axis of rotation of hoop passes through point B, we know that velocity of point O (centre of loop) is v. If ω is angular velocity of hoop then,

v = R ω (R is Distance between axis and centre(O))

Where ω is angular velocity.

i.e. ω = v/R = 0.2/2 = .1 s^{-1}

Moment of inertia of hoop, I = MR^{2}

Where, M is mass of hoop

R is Radius of hoop

∴ I = 100(2)^{2}

I = 400 Kg m^{2}

Total Kinetic Energy of hoop,

KE = KE_{rot} + KE_{tran}

KE_{rot} = 1/2 400 × (.1)^{2}

= 2 J

KE_{tran} = 1/2 100 × (.2)^{2}

= 2 J

∴ Total Kinetic energy,

KE = 2 + 2 = 4 J

Work done to stop the moving hoop is equal to its KE which is 4 J

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