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A disk of mass 2/3kg is rolling without slipping on a horizontal surface. The centre of the disk has a velocity of 4m/s. how much work is done to stop it?
Asked by shanusunny15 | 26 Oct, 2022, 20:46: PM
Expert Answer
Required workdone to stop the rolling disc equals its kinetic energy
Kinetic energy of rolling disc , E = (1/2) I ω2 + (1/2) m v2
where first term in RHS is rotational kinetic energy and second term is due to traslational kinetic energy .
I = (1/2) m R2 is moment of inertia , m is mass and R is radius ,
ω = ( v/R ) is angular velocity of rotation and v is velocity of centre of mass
E = (1/2) (1/2) m R2 ( v / R )2 + (1/2) m v2 = (3/4) m v2
E = (3/4) (2/3) 42 = 8 J
8 J energy is required to stop the disc
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