Question

A20 Kg solid disk /I=1/2Mr/ roll on a horizontal surface at the rate of 4m/s.compute its total kinetic energy

Answer 1

Explanation:

ke=1/2mv2+1/2I(angular velocity)2

Answer 2

Answer:

The total kinetic energy of the disk is equal to 240J which rolls on a horizontal surface.

Explanation:

Given, The mass of the disc will be equal to 20kg

The velocity with the disc rolls on a horizontal surface, $v=4ms^{-1}$

The total kinetic energy of the disc is given by the sum of the kinetic energy due to translational motion and rotational energy.

The kinetic energy due to translational motion:

$K.E.=\frac{1}{2} mv^{2}$

And, the rotational energy will be:

$K.E.=\frac{1}{2}I\omega ^{2}$

Where $I$ is the moment of inertia and ω is angular velocity.

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

where r is the radius of the disk.

Total kinetic energy: $K.E.=\frac{1}{2} mv^{2}+\frac{1}{2}I\omega ^{2}$

$K.E.=\frac{1}{2} mv^{2}+\frac{1}{2}(\frac{1}{2}mr^{2}\omega ^{2} )$

$K.E.=\frac{1}{2} mv^{2}+\frac{1}{2}(\frac{1}{2}m(r\omega )^{2} )$

where $v=r\omega$

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

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

Put the value of m and v in above equation:

$K.E.=\frac{3}{4}(20)(4)^{2} kgms^{-2}$

$K.E.=240J$

Therefore, the total kinetic energy will be equal to 240J.