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A solid sphere of mass 50 gm and diameter 2 cm rolls without sliding with a uniform velocity of 5 cm per sec, along a straight line of a smooth horizontal table, Its total energy is 


1. 87.5 × 10-6 joule
2. 25 × 10-6 joule
3. 250 × 10-6 joule
4. 62 × 10-6 joule

1 Answer

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Best answer
Correct Answer - Option 1 : 87.5 × 10-6 joule

CONCEPT:

Kinetic energy (K.E): 

  • The energy possessed by a body by virtue of its motion is called kinetic energy.
  • The expression for kinetic energy is

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

Where m = mass of the body and v = velocity of the body

Rotational Kinetic energy (KE):

  • The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
  • body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
  • Mathematically rotational kinetic energy can be written as

\(KE = \frac{1}{2}I{\omega ^2}\)

Where I = moment of inertia and ω = angular velocity.

CALCULATION: 

Given - m = 50 g = 0.05 kg, v = 5 cm/s = 0.05 m/s

  • The moment of inertia of the rolling ball is

\(\Rightarrow I=\frac{2}{5}mR^2\)

  • Kinetic energy due to the rotation is 

\(\Rightarrow K_2 = \frac{1}{2}I\omega^2\\ \Rightarrow K_2=\frac{1}{2}\times \frac{2}{5}mR^2\times\frac{v^2}{R^2}\\ \Rightarrow K_2=\frac{1}{5}mv^2\)

  • Total kinetic energy

\(\Rightarrow K=K_1+K_2\\ \Rightarrow K=\frac{1}{2}mv^2+\frac{1}{5}mv^2\\ \Rightarrow K=\frac{7}{10}mv^2\\\Rightarrow K =0.7\times 0.05\times (0.05)^2\\\Rightarrow K =87.5\times 10^{-6} J\)

  • Therefore, option 1 is correct.

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