Question:

A ring rolls down an inclined plane. The ratio of rotational kinetic energy to translational kinetic energy is

Updated On: May 12, 2024
  • 1:03
  • 1:01
  • 3:01
  • 2:01
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Translational kinetic energy,
KT=12mv2K_T = \frac{1}{2} mv^2
where mm = mass of the ring,
vv = speed of the centre of mass of the ring
Rotational kinetic energy,
KR=12Iω2K_R = \frac{1}{2} I \omega^2
Here, I=mR2,ω=vRI = mR^2 , \omega = \frac{v}{R}
So, KR=12(mR2)(vR)2=12mv2K_R = \frac{1}{2} ( mR^2 ) \left(\frac{v}{R} \right)^2 = \frac{1}{2} mv^2
    KRKT=11 \therefore\:\:\:\: \frac{K_R}{K_T} = \frac{1}{1}
Was this answer helpful?
0
0

Top Questions on System of Particles & Rotational Motion

View More Questions

Concepts Used:

System of Particles and Rotational Motion

  1. The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
  2. The distance between the pair of particles in such a body does not replace or alter. Rotational motion can be described as the motion of a rigid body originates in such a manner that all of its particles move in a circle about an axis with a common angular velocity.
  3. The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.