The coefficient of static friction between the tires of a motorcycle and the road is µs = 0.46. At a speed of?
The coefficient of static friction between the tires of a motorcycle and the road is µs = 0.46. At a speed of 13 m/s, what is the radius of the tightest unbanked turn that the driver can handle?
answer in m.
Cannot figure this dang question, please help.
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Tagged with: coefficient of static friction • motorcycle • radius • tires
Filed under: motorcycle tires
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I would take the time to do this but Steve hit the nail on the head! Good job Steve.
The formula for centripetal force is
F = mv^2 / r
The centripetal force here is .46 * m * g
So
.46 * g = v^2 / r
.46 * g = 13^2 / r
Amax = g*Cf = 9.8*.46 = 4.51 m/s²
R = V²/(Amax) = 13²/4.51 = 37.49 m