Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by v2 = 2gh / [1 + (k2/R2) ] Using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-7 System Of Particles And Rotational Motion



Q.27:- Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by v2 = 2gh / [1 + (k2/R2) ]
Using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

 

 

Answer:-

A body rolling on an inclined plane of height h,is shown in the following figure:



m = Mass of the body

R = Radius of the body
K = Radius of gyration of the body
v = Translational velocity of the body
h =Height of the inclined plane
g = Acceleration due to gravity
Total energy at the top of the plane, E­1= mgh
Total energy at the bottom of the plane, Eb = KErot + KEtrans
= (1/2) I ω2 + (1/2) mv2
But I = mk2 and ω = v / R
Eb = (1/2)(mk2)(v2/R2) + (1/2)mv2
= (1/2)mv2 (1 + k2 / R2)
From the law of conservation of energy, we have:
ET = Eb
mgh = (1/2)mv2 (1 + k2 / R2)
v = 2gh / (1 + k2 / R2)
Hence, the given result is proved.