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OLga [1]
3 years ago
8

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

ficient of static friction between tires and track is 0.37?
Physics
1 answer:
Aleks [24]3 years ago
3 0
First, let's put 22 km/h in m/s:

22 \frac{km}{h} \times  \frac{1000m}{1km}  \times  \frac{1h}{3600s}=6.11 \frac{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

F_{rad}=m \frac{v^2}{R}

The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:

F_{fric}= mg \mu_{s}

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:

m\frac{v^2}{R} = mg \mu_{s} \\ \\ \frac{v^2}{R} = g \mu_{s} \\ \\R= \frac{v^2}{g \mu_{s}} \\ \\ R= \frac{6.11}{9.8 \times 0.37}=1.685m

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miss Akunina [59]

Formula of the gravitational force between two particles:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67 \cdot 10^{-11} Nm^2 kg^{-2} is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance


(a) particle A

The gravitational force exerted by particle B on particle A is

F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N to the right

The gravitational force exerted by particle C on particle A is

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So the net gravitational force on particle A is

F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N to the right


(b) Particle B

The gravitational force exerted by particle A on particle B is

F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N to the left

The gravitational force exerted by particle C on particle B is

F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N to the right

So the net gravitational force on particle B is

F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N to the right


(c) Particle C

The gravitational force exerted by particle A on particle C is

F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N to the left

The gravitational force exerted by particle B on particle C is

F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N to the left

So the net gravitational force on particle C is

F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N to the left



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