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?
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 = KErot + KEtran
Where KErot is rotational kinetic energy and
KEtran is translational kinetic energy
KErot = 1/2 Iω2
KEtran = 1/2 Mv2
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 = MR2
Where, M is mass of hoop
R is Radius of hoop
∴ I = 100(2)2
I = 400 Kg m2
Total Kinetic Energy of hoop,
KE = KErot + KEtran
KErot = 1/2 400 × (.1)2
= 2 J
KEtran = 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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Separation of Motion of a system of particles into motion of the center of mass and motion about the center of mass:
Where pi is the momentum of the ith particle (of mass mi) and Note is the velocity of the ith particle relative to the canter of mass.
Also, prove using the definition of the centre of mass
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