A uniform disc of mass `1 kg` and radius `20 cm` is rolling purely on a flat horizontal surface. Its centre `C` is moving with acceleration `a = 20 ms^(-2)` and velocity `v =4 m//s` at a certain instant. At this instant. points `A` and `B` are located on the disc as shown in the diagram, with `AC = BC = 10 cm`.

What is the kinetic energy of the disc?

Options

A. `12J`
B. `8J`
C. `20J`
D. `10J`

Answer

A

Explanation

`KE=1/2mv^(2)=1/2Iomega^(2)=3/4mv^(2)=12J`
`veca_(A//C)=veca_(A)-veca_(C)`
` implies veca_(A)=veca_(A//C)+veca_(C)=omega^(2)rhati+alpharhatj+ahati=60hati+10hatj`
`veca_(A)=sqrt(3600+10)=10sqrt(37)m//s^(2)`
`veca_(B)=omega^(2)rhatj+alphar(-hati)+ahati=10hati+40hatj`
`a_(B)=10sqrt(17)m//s^(2)`