- #1
AN630078
- 242
- 25
- Homework Statement
- Hello, I have a question on the circular motion of the road but I am having a little difficulty answering it, specifically question 1 to find the radius from which other parts of the question lead on.
A cyclist is travelling on a bend in a road at 10ms^-1 and leans his bicycle at 20 degrees to the vertical in order to stay on the road. The total mass of the bicycle and cyclist is 90kg.
Question 1: What is the radius of the curvature of the bend of the road?
Question 2: On the same bend of the road a car of mass 1500kg travels at 25ms^-1. What frictional force would be required to keep the car’s tyres on the road?
Question 3: To eliminate the need for the frictional force on the cars tyres the road is banked, at what angle would this be?
I am very uncertain of my workings and wonder whether anyone may be of help, I would be very grateful for any guidance. I have attempted to solve the question as one can see but I do not know whether my approach would here be correct
- Relevant Equations
- centripetal acceleration=v^2/r
Centripetal force = mv^2/r
Force=mass*acceleration
θ=tan^-1(v^2/rg)
Question 1:
So we are given three variables;
Mass=90kg
Angle to the vertical = 20 degrees
Speed = 10 ms^-1
There is not enough information to rearrange the formulas for centripetal force or acceleration in terms of r to find the radius. However, I have a attached a free body diagram of a cyclist on a bend I found online.
The only external forces acting on the car are its weight w and the normal force of the road N, a frictionless surface can only exert a force perpendicular to the surface, a normal force. These two forces add to give a net external force that is horizontal toward the center of curvature and has magnitude mv2/r (the centripetal force). If the angle the cyclist leans θ is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. Moreover, only the Normal force has a horizontal component which must be equal to the centripetal force; Nsinθ=mv^2/r.
The net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. The vertical component of the normal force is N cos θ, and the other vertical force is the weight which must must be equal in magnitude; thus, N cos θ = mg.
Therefore, N=mg/cosθ
Substitute into the first equation;
mg(sinθ/cosθ)=mv^2/r
mg tan (θ)=mv^2/r
tan (θ)= v^2/rg
θ=tan^-1(v^2/rg)
(In hindsight I fear that in resolving the perpendicular component I have neglected the friction parallel to the road, which must supply the centripetal force. However, if the angle θ is ideal and at a certain speed there will not be a frictional force between the bicycle and the road. Is this formula correct or would it be wrong and not applicable here?)
However, if we rearrange this equation in terms of r;
r=v^2/g*tanθ
If I now input the known values to solve for r;
r=10^2/9.81*tan20
r=28.0069...~ 28.0 m to 3sf
Question 2;
As a car makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion.
F=mv^2/r
F=1500*25^2/28
F=33482.14286 ~ 33500 N to 3.s.f
Question 3:
Using the formula earlier derived;
θ=tan^-1(v^2/rg)
θ=tan^-1(25^2/28*9.81)
θ=66.27508...~ 66.3 degrees
So we are given three variables;
Mass=90kg
Angle to the vertical = 20 degrees
Speed = 10 ms^-1
There is not enough information to rearrange the formulas for centripetal force or acceleration in terms of r to find the radius. However, I have a attached a free body diagram of a cyclist on a bend I found online.
The only external forces acting on the car are its weight w and the normal force of the road N, a frictionless surface can only exert a force perpendicular to the surface, a normal force. These two forces add to give a net external force that is horizontal toward the center of curvature and has magnitude mv2/r (the centripetal force). If the angle the cyclist leans θ is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. Moreover, only the Normal force has a horizontal component which must be equal to the centripetal force; Nsinθ=mv^2/r.
The net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. The vertical component of the normal force is N cos θ, and the other vertical force is the weight which must must be equal in magnitude; thus, N cos θ = mg.
Therefore, N=mg/cosθ
Substitute into the first equation;
mg(sinθ/cosθ)=mv^2/r
mg tan (θ)=mv^2/r
tan (θ)= v^2/rg
θ=tan^-1(v^2/rg)
(In hindsight I fear that in resolving the perpendicular component I have neglected the friction parallel to the road, which must supply the centripetal force. However, if the angle θ is ideal and at a certain speed there will not be a frictional force between the bicycle and the road. Is this formula correct or would it be wrong and not applicable here?)
However, if we rearrange this equation in terms of r;
r=v^2/g*tanθ
If I now input the known values to solve for r;
r=10^2/9.81*tan20
r=28.0069...~ 28.0 m to 3sf
Question 2;
As a car makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion.
F=mv^2/r
F=1500*25^2/28
F=33482.14286 ~ 33500 N to 3.s.f
Question 3:
Using the formula earlier derived;
θ=tan^-1(v^2/rg)
θ=tan^-1(25^2/28*9.81)
θ=66.27508...~ 66.3 degrees