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?

A hoop of radius 2 m weights 100 kg. It rolls along a horizontal floor

1.	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?
Answer» Solution :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.