Question

A sphere of mass 0.5 kg and diameter 1m rolls without sliding with a constant velocity of 5 m/s. Calculate the ratio of the rotational kinetic energy to the total kinetic energy of the sphere

Choose the correct answer:

(a)7/10
(b)5/7
(c)2/7
(d)4/7​

Answer 1

Answer:

7/10 is correct answer

Answer 2

The ratio of the rotational kinetic energy to the total kinetic energy of the sphere is 2:7.

Explanation:

Given that,

Mass of the sphere, m = 0.5 kg

Diameter of the sphere, d = 1 m

Radius, r = 0.5 m

The sphere is sliding with a velocity of 5 m/s, v = 5 m/s

The rotational kinetic energy of an object is given by :

$E_r=\dfrac{1}{2}I\omega^2$........(1)

The kinetic energy of an object is given by :

$E_k=\dfrac{1}{2}mv^2$

Total energy of the object,

$T=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$.......(2)

the moment of inertia of sphere, $I=\dfrac{2}{5}mr^2$

Dividing equation (1) and (2) we get :

$\dfrac{E_r}{E_t}=\dfrac{\dfrac{1}{2}I\omega^2}{\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2}$

Since, $\omega=\dfrac{v}{r}$

$\dfrac{E_r}{E_t}=\dfrac{\dfrac{2}{5}\times mr^2\omega^2}{\dfrac{1}{2}mv^2+\dfrac{2}{5}\times mr^2\omega^2}$

$\dfrac{E_r}{E_t}=\dfrac{\dfrac{1}{2}\cdot\dfrac{2}{5}\cdot0.5^{2}\cdot\dfrac{1}{0.5^{2}}}{\dfrac{1}{2}\cdot\dfrac{2}{5}\cdot0.5^{2}\cdot\dfrac{1}{0.5^{2}}+\dfrac{1}{2}}$

moment of inertia of sphere,

$\dfrac{E_r}{E_t}=\dfrac{2}{7}$

So, the ratio of the rotational kinetic energy to the total kinetic energy of the sphere is 2:7. Hence, this is the required solution.                

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Kinetic energy

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