Question

A homogeneous cylinder of radius 20 cm and mass 45 kg is rolling without slipping along a horizontal floor at 2 m/s. How much work was required to give it this motion? Answer in units of kJ.

Answer 1

Answer:

$K = 0.135 KJ$

Explanation:

Total kinetic energy of a solid cylinder to perform pure rolling is given as

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

here we know that

$I = \frac{1}{2}mR^2$

$\omega = \frac{v}{R}$

now we have

$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v^2}{R^2})$

$K = \frac{1}{2}mv^2 + \frac{1}{4}mv^2$

so total kinetic energy will be

$K = \frac{3}{4}mv^2$

$K = \frac{3}{4}(45)(2^2)$

$K = 135 J$

$K = 0.135 KJ$

Answer 2

The work required to give this motion willbe equal to K= 0.135 Kj

What is work?

The work is defined as when force is applied on any object to displace it from its initial position to final position, then the product of the force to the displacement of the object is called as work done.

Total kinetic energy of a solid cylinder to perform pure rolling is given as

$K=\dfrac{1}{2}mv^2+\dfrac{1}{2}Iw^2$

here we know that

$I=\dfrac{1}{2}mR^2$

$w=\dfrac{v}{R}$

now we have

$K=\dfrac{1}{2}mv^2+\dfrac{1}{2}(\dfrac{1}{2}mR^2(\dfrac{v^2}{R^2})$

$K=\dfrac{1}{2}mv^2+\dfrac{1}{4}mv^2$

so total kinetic energy will be

$k=\dfrac{3}{4}mv^2$

$K=\dfrac{3}{4}(45)(2^2)$

$K=135\ J$

$K=0.135\ KJ$

Thus the work required to give this motion willbe equal to K= 0.135 Kj

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