Question

when a sphere of moment of inertia I moves down an inclined plane the percentage of energy which is rotational,is approximately
​

Answer 1

Answer:

Let us assume,

Mass of sphere = M

Radius of sphere = R

Moment of Inertia ($\rm I$) of a solid sphere passing through its centre of any diameter is given as

$\red{\implies \rm I = \dfrac{2}{5}MR^2}$

We know that

$\green{\implies \rm v = R\omega}$

$\green{\implies \rm \omega = \dfrac{v}{R}}$

Since,

⇒ Total energy = Rolling energy + Transnational energy

$\blue{\implies \rm Total \;Energy = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}Mv^2}$

$\blue{\implies \rm T.E = \dfrac{1}{2} \times \dfrac{2}{5}MR^2 \times \dfrac{v^2}{R^2} + \dfrac{1}{2}Mv^2}$

On simplification,

We get,

$\implies \rm Total \;Energy = \dfrac{7}{10}Mv^2$

We are asked to calculate the % of Rotational Kinetic Energy

Therefore,

Dividing it by Total Energy We get

$\purple{\implies \rm \% \; of \;Rotational\;Energy = \dfrac{\dfrac{1}{5}Mv^2 }{\dfrac{7}{10} Mv^2} \times 100}$

$\pink{\implies \rm \% \; of \;Rotational\;Energy = 28.5667 \;\%}$

$\orange{\implies \rm \% \; of \;Rotational\;Energy \approx 28\;\%}$

Answer is 28 %