Question

For the following objects, which all roll without slip- ping, determine the rotational kinetic energy about the center of mass as a percentage of the total kinetic energy: (a) a solid sphere, (b) a thin spherical shell, and (c) a tiin cylindrical shell.

Answer 1

Answer:

The answers to the question are

(a) 28.57 %

(b) 40 %

(c) 50 %

Explanation:

The following are the relations between total and rotational kinetic energies of a sphere

$KE_{Total} = KE_{Translational} +KE_{Rotational}$ =

$KE_{Total} =\frac{1}{2} m v^{2} + \frac{1}{2} I\omega ^{2}$

ω = angular velocity

v = velocity

m = mass

I = Moment of inertia = $\frac{2}{5} *m*r^{2}$  and ω = v/r

Therefore $KE_{Total} =\frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} )*m*r^{2}*\frac{v}{r} ^{2}$ = $KE_{Total} =\frac{1}{2} m v^{2} + \frac{1}{5} mv^{2} = \frac{7}{10} mv^{2}$

Therefore $\frac{KE_{Total} }{KE_{Rotational} } Percent= \frac{\frac{1}{5} mv^{2} }{\frac{7}{10}mv^{2} } *100 = \frac{10}{35} *100 = 28.57 %$

= 28.57 %

(b) for a thin sherical shell, I = $\frac{2}{3} mr^{2}$ Therefore $KE_{Rotational} = \frac{1}{2} I\omega^{2}$

= $(\frac{1}{2}) (\frac{2}{3}) mr^{2} *\omega ^{2}$ =$(\frac{1}{3}) I*(\frac{v}{r} )^{2}$ = 1/3mv²

KE(Total) = KE(Transitional) + KE(Rotational) = 1/2mv² + 1/3mv² = 5/6mv²

KE(Transitional) /KE(Total) = (1/3mv²)/(5/6mv²) = 6/15

Percentage total = 6*100/15 = 40 %

(c) Thin cylindrical shell = I = $mr^{2}$ and $KE_{Rotational} = \frac{1}{2} I\omega^{2} = \frac{1}{2} mr^{2}(\frac{v}{r}) ^{2} = \frac{1}{2} mv^{2}$

Therefore $KE_{Total} = KE_{Translational} +KE_{Rotational}$

$KE_{Total} =\frac{1}{2} m v^{2} + \frac{1}{2} mv ^{2} =mv ^{2}$ and $\frac{KE_{Rotational}}{KE_{Total} } = \frac{ \frac{1}{2} m v^{2}}{mv ^{2} } =\frac{1}{2}$

In percentage = $\frac{1}{2} * 100$ = 50 %