Question

A ring rolls on a plane surface. The fraction of total
energy associated with its rotation is :
(a)1/2
(b) 1
(c)1/4
(d)2/1


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Answer 1

Answer:

In case of rolling without slipping the angular velocity (ω) and the velocity of centre of mass (v) are related as: v = Rω.

>>Kr = 1/3

Explanation:

Given:

A ring rolls on A plane surface.the fraction of its total energy associated with its rotation is

Moment of inertia of the Ring (disc), I = 1/2mr2

Solution:

Total kinetic energy, K = Translational kinetic energy + Rotational kinetic energy

or, K = 1/2 mv2 + 1/2 Iω2

We can write rotational kinetic energy as: 

1/2 Iω2 = (1/2 )×(1/2 mr2) (ω2) = 1/4 m (rω)2 = 1/4 mv2 

Thus,

K total = Kt + Kr 

2Kr = Kt

Ktotal = 3 Kr

Fraction of rotational kinetic energy

Kr/3 Kr = 1/3

Answer 2

THE REQUIRED FRACTION IS 1/2.

Given:

• Ring rolls on plane surface.

To find:

• Fraction of energy associated with rotation.

Calculation:

First of all, the total energy for any rolling object is given as :

$TE = \dfrac{1}{2} m {v}^{2} (1 + \dfrac{ {k}^{2} }{ {r}^{2} } )$

• m is mass
• v is velocity
• k is radius of gyration
• r is radius

Now, kinetic energy for rotation is :

$KE_{r}= \dfrac{1}{2} m {v}^{2}( \dfrac{ {k}^{2} }{ {r}^{2} } )$

Now, taking appropriate ratios:

$\dfrac{KE_{r}}{ TE} = \dfrac{ \dfrac{1}{2}m {v}^{2}( \dfrac{ {k}^{2} }{ {r}^{2} } ) }{\dfrac{1}{2} m {v}^{2} (1 + \dfrac{ {k}^{2} }{ {r}^{2} } )}$

$\implies \dfrac{KE_{r}}{ TE} = \dfrac{ \dfrac{ {k}^{2} }{ {r}^{2} } }{1 + \dfrac{ {k}^{2} }{ {r}^{2} } }$

• k²/r² for a ring is 1.

$\implies \dfrac{KE_{r}}{ TE} = \dfrac{ 1 }{1 + 1 }$

$\implies \dfrac{KE_{r}}{ TE} = \dfrac{ 1 }{ 2 }$

So, the fraction is ½ (Option a)