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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 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 - Physics

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 `v^2 = (2gh)/((1+k^2"/"R^2))`.

Using dynamical consideration (i.e. by consideration of forces and torques). Note 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.

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Solution 1

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

m = Mass of the body

= Radius of the body

K = Radius of gyration of the body

= Translational velocity of the body

=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, `E_b = KE_rot + KE_trans`

`=1/2 Iomega^2 + 1/2 mv^2`

But `I = mk^2 " and " omega = v/r`

`:.E_b = 1/2 (mk^2)(v^2/R_2) + 1/2 mv^2`

`=1/2 mv^2 k^2/R^2 + 1/2mv^2`

`= 1/2 mv^2(1+ k^2/R^2)`

From the law of conservation of energy, we have:

`E_T = E_b`

`mgh = 1/2mv^2(1+k^2/R^2)`

`:.v = (2gh)/(1+k^2"/"R^2)`

Hence, the given result is proved.

Solution 2

Let a rolling body (I = Mk2) rolls down an inclined plane with an initial velocity u = 0; When it reaches the bottom of the inclined plane, let its linear velocity be v. Then from conservation of mechanical energy, we have Loss in P.E. = Gain in translational K.E. + Gain in rotational K.E.

`Mgh = 1/2mv^2 + 1/2 Iomega^2`

`= 1/2mv^2 + 1/2(mk^2)(v^2/R^2)`

`Mgh = 1/2mv^2 (1+k^2/R^2)`

`v^2 = (2gh)/(1+k^2/R^2)`

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NCERT Class 11 Physics Textbook
Chapter 7 System of Particles and Rotational Motion
Q 27 | Page 180
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