Question #b1c2e

1 Answer
A08
Jan 15, 2017

(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

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