A motor cyclist moving with a velocity of $72 ext{ km/hour}$ on a flat road takes a turn on the road at a po

Question

A motor cyclist moving with a velocity of $72 	ext{ km/hour}$ on a flat road takes a turn on the road at a point where the radius of curvature of the road is $20 	ext{ meters}$. The acceleration due to gravity is $10 	ext{ m/sec}^2$. In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater than:

Accepted Answer

1. **Convert Units:** First, convert the velocity from km/hr to m/s.
   $$ v = 72 	imes \frac{5}{18} = 20 	ext{ m/s} $$
2. **Identify the Physics Principle:** When a cyclist takes a turn, they lean inward to generate the necessary centripetal force from the horizontal component of the normal reaction. The angle $	heta$ with the vertical is given by the relationship:
   $$ 	an 	heta = \frac{v^2}{rg} $$
   [Source 41]
3. **Substitute Values:**
   - Velocity $v = 20 	ext{ m/s}$
   - Radius $r = 20 	ext{ m}$
   - Gravity $g = 10 	ext{ m/s}^2$
   $$ 	an 	heta = \frac{(20)^2}{20 	imes 10} $$
   $$ 	an 	heta = \frac{400}{200} = 2 $$
4. **Solve for Angle:**
   $$ 	heta = 	an^{-1}(2) $$
   (Note: Option 3 corresponds to the error of failing to convert km/hr to m/s: $72^2/200 = 25.92$).

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