Question

Question #b1c2e

Answer 1

(b)

Explanation:

When disc of mass `m` and radius `R` rolls down a hill with a velocity `v` of its centre of mass, without slipping, it has three types of energies associated with it at any instant of time

1. Potential energy `PE=mgh`
2. Rotational Kinetic Energy given by `KE_R=1/2Iomega^2`
   where moment of inertia `I=1/2mR^2` and angular velocity `omega=v/R`
3. Translational Kinetic energy `KE_T=1/2mv^2`

(We have ignored friction in this statement.)

We also know that

(a) Initial energy at the top of hill when disc is at rest it has only `PE`
(b) At the bottom of hill all the `PE` gets converted into sum of `KE_T` and `KE_R`

By Law of Conservation of energy

`TE_"(a)"=TE_"(b)"`

`mgh=1/2Iomega^2+1/2mv^2`
`=>mgh=1/2(1/2mR^2)(v/R)^2+1/2mv^2`
`=>gh=1/4v^2+1/2v^2`
`=>gh=3/4v^2`
`=>v=sqrt(4/3gh)`

Inserting given values we get
`v=sqrt(4/3xx 9.81xx10)`
`vapprox11.43ms^-1`