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A daredevil drives a motorcycle around a circular vertical wall 100 ft in diameter. The coefficient of friction between tires and the wall is 0.60. At what angle will the motorcycle be inclined to the horizontal? What is the effect of traveling at a greater speed?
A.) V= 35.3 mph, angle= 31°
B.) V= 56.2 mph, angle= 56°
C.) V= 12 mph, angle= 34°
show solution step by step
thanks
A.) V= 35.3 mph, angle= 31°
B.) V= 56.2 mph, angle= 56°
C.) V= 12 mph, angle= 34°
show solution step by step
thanks
Answers
Answered by
Marian
A) V=35.3 mph, angle=31 deg
g=gravity, u=coefficient of friction, r=radius, m=mass (which will be cancelled out later)
N=m(v^2/r)
uN=mg
u(m(v^2/r)=mg
You will get a formula of √rg/n for velocity
v=√50(32.2)/0.6
v=51.8 ft/s or 35.3 mph
g=gravity, u=coefficient of friction, r=radius, m=mass (which will be cancelled out later)
N=m(v^2/r)
uN=mg
u(m(v^2/r)=mg
You will get a formula of √rg/n for velocity
v=√50(32.2)/0.6
v=51.8 ft/s or 35.3 mph
Answered by
Marian
For the angle..
the formula of coefficient of friction is tan θ=f
θ=tan^-1(0.60)
θ=30.96º >>degree of friction
the formula to use for the angle from the horizontal is
tan(θ(horizontal)-θ(friction))=v²/gr
input the values and you will get
θ=329.034485º / 31º from the horizontal
the formula of coefficient of friction is tan θ=f
θ=tan^-1(0.60)
θ=30.96º >>degree of friction
the formula to use for the angle from the horizontal is
tan(θ(horizontal)-θ(friction))=v²/gr
input the values and you will get
θ=329.034485º / 31º from the horizontal
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