Question

when a disc is rolling along level surface of percentage of its total kinetic energy that is transitional is nearly
a. 33%
b. 66%
c. 50%
d. 25%​

Answer 1

Given:

A disc is rolling along level surface.

To find:

Percentage of its total kinetic energy that is transitional is nearly ?

Calculation:

Let velocity of centre of mass of the disc be "v":

Translational Kinetic Energy in rolling motion:

$\rm \therefore \: KE_{(translational)} = \dfrac{1}{2} m {v}^{2}$

Total Kinetic Energy in rolling motion :

• "k" is radius of gyration.

$\rm \therefore \: KE_{(total)} = \dfrac{1}{2} m {v}^{2} \bigg(1 + \dfrac{ {k}^{2} }{ {r}^{2} } \bigg)$

• For a disc , value of k²/r² = ½

$\rm \implies\: KE_{(total)} = \dfrac{1}{2} m {v}^{2} \bigg(1 + \dfrac{1 }{ 2 } \bigg)$

$\rm \implies\: KE_{(total)} = \dfrac{1}{2} m {v}^{2} \bigg( \dfrac{3}{ 2 } \bigg)$

So, fraction of total KE , which is translational:

$\rm \implies\: \dfrac{KE_{(translational)}}{KE_{(total)}} = \dfrac{ \frac{1}{2} m {v}^{2} }{\frac{1}{2} m {v}^{2} \bigg( \frac{3}{ 2 } \bigg)}$

$\rm \implies\: \dfrac{KE_{(translational)}}{KE_{(total)}} = \dfrac{2}{3}$

$\rm \implies\: \bigg(\dfrac{KE_{(translational)}}{KE_{(total)}} \bigg)\%= \dfrac{2}{3} \times 100\%$

$\rm \implies\: \bigg(\dfrac{KE_{(translational)}}{KE_{(total)}} \bigg)\% \approx 66\%$

So, the fraction of total KE , which is translational is 66%.