What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 22 km/h and the coef
1 answer:
First, let's put 22 km/h in m/s:
Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by
The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:
Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:
Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by
The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:
Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:
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Answer:
a) L = 3.29 10⁻⁴ H, b)U = 5.33 10⁻² J
Explanation:
a) The inductance is a solenoid this given carrier
L =
The magnetic field inside the solenoid is
B = μ₀
hence the magnetic flux
Ф_B = B. A = μ₀
we substitute in the expression of inductance
L = N² μ₀ A /l
let's find the area of each turn
A = π r²
A = π 0.02²
A = 1.2566 10⁻³ m²
let's calculate
L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3
L = 3.29 10⁻⁴ H
b) The stored energy is
U = ½ L i²
let's calculate
U = ½ 3.29 10⁻⁴ 18²
U = 5.33 10⁻² J