Question

A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base of the hill. (a) At what rate was it rotating at the base of the hill? (b) What was the rotational kinetic energy then? (Ignore rolling friction and assume total mechanical energy is conserved). [Hint: Soccer ball is a thin walled h

Answer 1

Answer with Explanation:

We are given that

Diameter=d=22.6 cm

Mass,m=426 g=$426\times 10^{-3} kg$

1 kg=1000 g

Radius,r=$\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m$

1m=100 cm

Height,h=5m

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

a.By law of conservation of energy

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

$\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh$

$v=\omega r$

$gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2$

$\omega^2=\frac{6}{5r^2}gh$

$\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s$

Where $g=9.8m/s^2$

b.Rotational kinetic energy=$\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J$

Rotational kinetic energy=8.35 J