Forces on a Motorcycle on a Vertical Circular Track

Problem Statement

A 250 kg motorcycle is driven around a 12-meter tall vertical circular track at a constant speed of $11 \, \mathrm{m/s}$.

Find:

1. The normal and friction forces at four points on the track:
   • (i) at the bottom (and rising)
   • (ii) halfway to the top
   • (iii) at the top
   • (iv) $45^{\circ}$ past the top
2. The minimum coefficient of static friction needed to complete the stunt as planned.

Assumptions: The mass of the motorcycle includes the mass of the rider. Aerodynamic drag and rolling resistance are negligible.

Given Data

• Mass of motorcycle and rider, $m = 250 \, \mathrm{kg}$
• Track height, $H = 12 \, \mathrm{m}$
• Speed, $v = 11 \, \mathrm{m/s}$
• Radius of the track, $R = H/2 = 6 \, \mathrm{m}$

Solution

(a) Normal and Friction Forces at Key Points

(i) At the Bottom (Rising)

• Radial direction: Let the $+r$ direction point towards the center of the track.

\[\sum F_r = F_n - F_g = m \frac{v^2}{R}\]

Rearranging, we get:

\[F_n = m \frac{v^2}{R} + mg\]

• Tangential direction: Let the $+x$ direction point left.

\[\sum F_x = -F_f = m \cdot 0 \Rightarrow F_f = 0\]

Since the motorcycle moves at constant speed, $a_x = 0$, so $F_f = 0$.

(ii) Halfway to the Top

• Radial direction: $+r$ direction points towards the center.

\[\sum F_r = F_n = m \frac{v^2}{R}\]

Thus:

\[F_n = m \frac{v^2}{R}\]

• Vertical direction: Let the $+y$ direction point upwards.

\[\sum F_y = F_f - F_g = m \cdot 0 \Rightarrow F_f = mg\]

Since $a_y = 0$ at constant speed, $F_f = mg$.

(iii) At the Top

• Radial direction: $+r$ direction points towards the center.

\[\sum F_r = F_n + F_g = m \frac{v^2}{R}\]

Solving for $F_n$:

\[F_n = m \frac{v^2}{R} - mg\]

• Tangential direction: Let the $+x$ direction point right.

\[\sum F_x = -F_f = m \cdot 0 \Rightarrow F_f = 0\]

Since $a_x = 0$, $F_f = 0$.

(iv) $45^{\circ}$ Past the Top

• Radial direction: $+r$ direction points towards the center.

\[\sum F_r = F_n + F_g \cos(45^{\circ}) = m \frac{v^2}{R}\]

Thus:

\[F_n = m \frac{v^2}{R} - mg \cos(45^{\circ})\]

• Tangential direction: $+x$ direction points tangentially down at $45^{\circ}$.

\[\sum F_x = -F_f + F_g \sin(45^{\circ}) = m \cdot 0 \Rightarrow F_f = mg \sin(45^{\circ})\]

Since $a_x = 0$, $F_f = mg \sin(45^{\circ})$.

(b) Minimum Coefficient of Static Friction

From case (ii), we require $F_f \geq mg$ for the motorcycle to stay on the track. Since $F_f \leq \mu F_n$ and $F_n = m \frac{v^2}{R}$ (from case ii), we have:

\[mg \leq F_f \leq \mu m \frac{v^2}{R}\]

Solving for $\mu$:

\[\mu \geq \frac{mg}{m \frac{v^2}{R}} = \frac{Rg}{v^2}\]

Thus, the minimum coefficient of static friction is:

\[\mu \geq \frac{Rg}{v^2}\]