Question

A solid cylinder of mass m is rolling down the inclined plane without slipping. What will be total kinetic energy of the mass if at any time the linear velocity of the mass is v?
1. \(\frac{1}{2}m{{v}^{2}}\)
2. \(\frac{2}{5}m{{v}^{2}}\)
3. \(\frac{3}{4}m{{v}^{2}}\)
4. ​\(m{{v}^{2}}\)​

Answer 1

Correct Answer - Option 3 : \(\frac{3}{4}m{{v}^{2}}\)

CONCEPT:

• The kinetic energy of the rotating body is given by

\(KE = \frac{1}{2}mv^2[1+\frac{K^2}{R^2}]\)

where m is the mass of the body, v is the velocity, and K/R is the ratio of the radius of gyration to the radius of the body.

Values to remember:

• \(\frac{K^2}{R^2} \) for the solid sphere is 2/5.
• \(\frac{K^2}{R^2} \) for the spherical shell is 2/3.
• \(\frac{K^2}{R^2} \) for the solid cylinder is 1/2.​

EXPLANATION:​

The kinetic energy of the rotating body is given by:

\(KE = \frac{1}{2}mv^2[1+\frac{K^2}{R^2}]\)

\(\frac{K^2}{R^2} \) for the solid cylinder is 1/2.

\(KE = \frac{1}{2}mv^2[1+\frac{1}{2}]\)

KE = \(\frac{3}{4}m{{v}^{2}}\)

So the correct answer is option 3.