Answer to Question #112610 in Classical Mechanics for Julie Benson

Question

A solid rolls over the top of a hill on a track, its speed is 80cm/sec. If friction losses are negligible, how fast is the disc moving when it is 80cm below the top?

Answer 1

According to the law of conservation of energy, the mechanical energy at the top will increase the speed of the disk when it is 80 cm below the top. Since the disk is rolling, we must also include the kinetic energy of rotation, which is

$$KE_R=\frac{1}{2}I\omega^2=\frac{1}{2}\cdot\frac{1}{2}mr^2\bigg(\frac{v}{r}\bigg)^2.$$

Therefore:

$$KE_{L.top}+KE_{R.top}+PE_{top}=KE_{L.bot}+KE_{R.bot}.$$

$$\frac{1}{2}mv^2+\frac{1}{4}mr^2\frac{v^2}{r^2}+mgh=\\ \space\\=\frac{1}{2}mu^2+\frac{1}{4}mr^2\frac{u^2}{r^2},\\ \space\\ \frac{3}{4}v^2+gh=\frac{3}{4}u^2,\\ \space\\ u=\sqrt{v^2+\frac{4}{3}gh}=\sqrt{0.8^2+\frac{4}{3}\cdot9.8\cdot0.8}=3.33\text{ m/s}.$$