The Kinetic Energy of Rolling — Question 8

Question 8

Residual spin limits the height reached on a smooth slope

A disk with $I=MR^2/2$ enters a frictionless incline at center speed $3\,\mathrm{m/s}$ and angular speed $6/R\,\mathrm{rad/s}$, where $R$ is in metres. No impact or impulsive torque occurs at entry.

1. Find the maximum height gained and the spin energy remaining there.

2. Explain why using all the initial kinetic energy to predict height fails.

Question 8 - Solution

Residual spin limits the height reached on a smooth slope

(a) Derivation and calculation

The smooth contact normal and gravity exert no torque about the center: $$\tau_C=0\Rightarrow\omega_f=\omega_i=6/R.$$ At maximum height, center translation stops: $$Mgh=\tfrac12M(3)^2\Rightarrow\boxed{h=\frac9{2g}}.$$ The remaining rotational energy is $$\boxed{K_{{\mathrm{rot}},f}=\tfrac12\left(\frac{MR^2}{2}\right)(6/R)^2=9M\ \mathrm{J/kg}}.$$

(b) Result and interpretation

The complete energy balance is $$\underbrace{4.5M}_{K_{{\mathrm{trans}},i}}+\underbrace{9M}_{K_{{\mathrm{rot}},i}} =\underbrace{Mgh}_{4.5M}+\underbrace{9M}_{K_{{\mathrm{rot}},f}},$$ with each numerical coefficient in J/kg. Converting all $13.5M$ of initial energy into height would violate the conserved spin. No rolling constraint exists on the frictionless slope.

Physical check

The initial spin is twice the value needed for ground rolling at 3 m/s, but this does not affect the smooth-ramp rise.