A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (Kt) as well as rotational kinetic energy (Kr) simultaneously. The ratio Kt ∶ (Kt + Kr) for the sphere is

  1. 7 ∶ 10
  2. 5 ∶ 7
  3. 2 ∶ 5
  4. 10 ∶ 7

Answer (Detailed Solution Below)

Option 2 : 5 ∶ 7
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Detailed Solution

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Concept:

Whenever object is in rolling motion then it possess two types of kinetic energy which are

1. Translational kinetic energy is due to linear motion of an object.

KT = \(\frac{1}{2}\)mv2

where m and v is mass and velocity of the object respectively.

2. Rotational kinetic energy due to rotational motion of an object.

KR = \(\frac{1}{2}\)I\(w\)2

where I and \(w\) is moment of Inertia and  angular velocity of the object respectively.

Thus, total kinetic energy of the object is

Total K.E. = KT +KR =  \(\frac{1}{2}\)mv + \(\frac{1}{2}\)I\(w\)2

Calculation:

If object is rolling, then the rotational kinetic energy is-

\({k_t} = \frac{1}{2}m{v^2}\)

\({k_t} + {k_r} = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2} = \frac{1}{2}m{v^2} + \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\left( {\frac{v}{r}}\right)^2}\)

\( = \frac{7}{{10}}m{v^2}\)

\(So,\frac{{{k_t}}}{{{k_t} + {k_r}}} = \frac{5}{7}\)

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