Question

A wheel of radius a, with its mass concentrated on the rim, is rolling with velocity v round a circle of radius (R >> a), maintaining a constant inclination alpha to the vertical. Show that v = a omega = R Ohm, where omega is the angular velocity of the wheel about its axis, and Ohm (<< Omega) is the precessional angular velocity of the axis. Use the momentum equation to find the horizontal and vertical components of the force at the point of contact. Then show from the angular momentum equation about the centre of mass that R = 2v^2/g tan alpha. Evaluate R for v = 5ms^-1 and alpha = 30 degree.

?explanations needed, thanks.

Answer 1

using equations from centripetal and centrifugal forces are

entripetal force = mv² / R

centrifugal force = mv² / R

and the total force is F

F = mv² / R + mv² / R

F = 2mv² / R ---- 1

and the force is with an angle $\alpha$

F = m g sin $\alpha$ / cos $\alpha$

F = m g tan $\alpha$ ----- 2

equating equations 1 and 2

m g tan $\alpha$ = 2mv² / R

g tan $\alpha$ = 2 v² / R

R = 2 v² / g tan $\alpha$

here given $\alpha$ = 30⁰ , v = 5 m/s

and g = 9.8 m/s²

then substituting values in the above equation

R = 2 X 5 X 5 / 9.8 X tan30

R = 50 / 9.8 X 0.577

R = 50 / 5.658

R = 8.837 m