Question #b1c2e

1 Answer
Jan 15, 2017

Answer:

(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#