A hoop of radius 2 m weights 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 stope it?

Explanation

Angular velocity of centre of mass of hoop
`omega=(v)/(r )=(0.20)/(2)`
`therefore omega=0.1" rad s"^(-1)`
Suppose moment of inertia about an axis passing through its centre and perpendicular to its plane is I.
`therefore I=Mr^(2)` (ring type)
= `100xx(2)^(2)`
`=400kgm^(2)`
Now initial total kinetic energy of hoop = rotational kinetic energy + translational kinetic energy
`K_(0)=(1)/(2)Iomega^(2)+(1)/(2)mv^(2)`
`=(1)/(2)xx400xx0.01+(1)/(2)xx100xx400xx10^(-4)`
`=2+2=4J`
From work energy theorem in
`W=K-K_(0),K=0`
`=-4J`
Negative sign indicate that work is done on hoop.