Problem — Acceleration of a wheel on an inclined plane

Problem

A wheel shaped like a solid cylinder ($I = \tfrac{1}{2}MR^2$) rolls without slipping down a $30^\circ$ inclined plane. Calculate the acceleration of the wheel’s centre of mass.

Solution

For a body that rolls without slipping, the acceleration along the plane is $a = \frac{g\sin\theta}{1 + I/(MR^2)}$ For the solid cylinder $I/(MR^2) = \tfrac{1}{2}$, so the denominator equals $1 + \tfrac{1}{2} = \tfrac{3}{2}$: $a = \frac{g\sin 30^\circ}{3/2} = \frac{9{,}81\cdot 0{,}5}{1{,}5}$ $\ev{a \approx 3{,}3\;\mathrm{m/s^2}}$ This is less than $g\sin\theta = 4{,}9\;\mathrm{m/s^2}$ (a block sliding without rotating): part of the effect of gravity goes into making the wheel spin.

Links

Topics: Dinamica rotazionale Concepts: Rotolamento · Accelerazione angolare Objects: Piano inclinato · Disco rotante