Question

A cylinder of mass 6.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 7.0 m/s. (a) Determine the translational kinetic energy of its center of mass. J (b) Determine the rotational kinetic energy about its center of mass. J (c) Determine its total energy. J

Answer 1

Answer:

(a) 147 j (b) 73.5 J (C) 220.5 J

Explanation:

Mass of the cylinder m =6 kg

velocity = 7 m/sec

(a) transnational energy $KE =\frac{1}{2}mv^2=0.5\times 6\times 7^2=147 \ j$

(b) Rotational kinetic energy is given by $KE =\frac{1}{2}I\omega ^2$ where I is moment of inertia and ω is the angular velocity

Moment of inertia $I=\frac{1}{2}mr^2$

Putting the value of I in rotational kinetic energy formula

$KE=\frac{1}{2}\frac{1}{2}mr^2\omega ^2$

$KE=\frac{1}{4}m(rw)^2$

$KE=\frac{1}{4}m(v)^2$ as v=ωr

So rotational kinetic energy $=\frac{1}{2}\times translational\ kinetic \ energy=0.5\times 147=73.5j$

(c) Total energy = 147+73.5=220.5 j