Question

A 2.50-kg solid, uniform disk rolls without slipping across a level surface, translating at 3.75 m/s. If the disk’s radius is 0.100 m, find its (a) translational kinetic energy and (b) rotational kinetic energy.

Answer 1

Answer

given,

mass of the solid = 2.50 Kg

speed of translating disk = 3.75 m/s

radius of the disk = 0.1 m

a) transnational Kinetic energy

    $KE = \dfrac{1}{2}mv^2$

    $KE = \dfrac{1}{2}\times 2.5 \times 3.75^2$

    $KE =17.58\ J$

b)  rotational kinetic energy.

    $KE = \dfrac{1}{2}I\omega^2$

     for disk

    $I = \dfrac{1}{2}mr^2$       and v = rω

    $KE = \dfrac{1}{2}( \dfrac{1}{2}mr^2)(\dfrac{v}{r})^2$

    $KE = \dfrac{1}{4}mv^2$

    $KE = \dfrac{1}{4}\times 2.5 \times 3.75^2$

    $KE =8.79\ J$