Question

A coin of radius b and mass M rolls on a horizontal surface at speed V. If the plane of the coin is vertical the coin rolls in a straight line. If the plane is tilted, the path of the coin is a circle of radius R. Find an expression for the tilt angle of the coin a in terms of the given quantities.(Because of the tilt of the coin the circle traced by its center of mass is slightly smaller than R but you can ignore the difference.)

Answer 1

The tilt angle is α=tan⁻¹(v²/2Rg)

Step-by-step explanation:

As the coin rolls with speed v around the circle of radius R, it rotates around the vertical at rate

Ω = v/R.

The rotation is caused by the precession of its spin angular angular momentum due to the torque caused by the tilt. For rolling without slipping,

v = bωₙ,

where

• ωₙ is the angular frequency of the coin
• b is the radius of the coin

so that

Ω = ωₙ(b/R)

The coin is accelerating, so take the torque about the centre of mass. From the force diagram:

τ_cm = fb×cos(α) - Nb×sin(α)

where

• f is the centrifugal force which is equal to Mv²/R

• N is the normal force which is equal to Mg

Therefore, the equation of motion for Lₙ is

τ_cm = ΩLₙ×cos(α)

= ΩI₀ωₙ×cos(α)

= ωₙ²×(b/R)×I₀cos(α)

= (v/b)²(b/R)(1/2(Mb²)cos(α)

= 1/2(Mv²)(b/R)cos(α)

= Mv²(b/R)cos(α) - Mg(b)sin(α)

Thus, the tilt angle is

tan(α) = (v²/2Rg) ⇒ α=tan⁻¹(v²/2Rg)