Question

a spherical ball rolls on a table without slipping determine the percentage of its kinetic energy which is rotational.moment of inertia of sphere2/5×mass×radius^2​

Answer 1

Answer:

$28.57\%$

Explanation:

Let the mass and the radius of the sphere are m and r respectively

Moment of Inertia of the sphere

$I=\frac{2}{5}mr^2$

Kinetic Energy of the rolling sphere

$=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$=\frac{1}{2}mv^2+\frac{1}{2}\times \frac{2}{5}mr^2\times \omega^2$

$=\frac{1}{2}mv^2+\frac{1}{2}\times \frac{2}{5}m\times (r\omega)^2$

$=\frac{1}{2}mv^2+\frac{1}{5}mv^2$

$=\frac{7}{10}mv^2$

Rotational Kinetic energy

$=\frac{1}{5}mv^2$

$\frac{\text{Rotational Kinetic Energy}}{\text{Total Kinetic Energy}}$

$=\frac{1}{5}mv^2/\frac{7}{10}mv^2$

$=\frac{2}{7}$

Percentage

$=\frac{2}{7}\times 100$

$=28.57\%$

Hope this helps.

Answer 2

Answer:

The percentage of rolling ball kinetic energy is 28.57%.

Given Data:

M.I of Sphere $=\frac{2}{5} \times \text { mass } \times \text { radius }^{2}.$

Explanation:

Step 1:

Total Kinetic energy of rolling sphere $=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$

But moment of inertia of sphere $I=\frac{2}{5} m r^{2}$

Step 2:

Therefore $\mathrm{K.E.}=\frac{1}{2} m v^{2}+\frac{1}{2} \times \frac{2}{5} m r^{2} \times \omega^{2}$

$K . E .=\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}$

$K . E .=\frac{7}{10} m v^{2}$

Step 3:

Now, the % of rolling $\mathrm{K.E.}=\frac{\frac{1}{5} m v^{2}}{\frac{7}{5} m v^{2}} \times 100$

% of rolling $K.E =\frac{200}{7}$

% of rolling $K.E = 28.57%$