First off, thanks a lot for your help.
Secondly, will you do it again?
A bicyclist traveling with speed v = 3.2m/s on a flat toad is making a turn with radius r = 2.8 m. The forces acting on the cyclist and cycle are the normal force (N) and friction force (Ffr) exerted by the road on the tires and mg, the total weight of the cyclist and bicycle.
A) Explain why the angle theta the bicycle makes with the verticle must be given by tan(theta) = Ffr/N if the cyclist is to remain balanced.
B) Find theta for the values given.
C) If the coefficient of static friction between the tires and road is µs = 0.65, what is the minimum turning radius?
I got the centripetal acceleration X mass = Ffr.
I got mg equals to N(hypothenuse * sin(th)) - Ffr(hypotenuse * cos(th))
In order to find the theta, I need to use circular translational motion?
This is where I got stumped.
A) Draw yourself a head-on view of the tilted bike coming out of the page. The torque about the point where the wheel touches the road must be zero, otherwise the bike will tip one way or the other. There is an upward force N on the tire that equals the weight of bike and rider. The opposite force acts as if applied to the center of mass of the bike/rider combination. There is also a centripetal friction force Ff applied at the point where the wheel meets the road. The problem can be treated statically if one allows a centrifugal or "D'Alembert" force to be acting horizontally on the bike at the center of mass of the bike-rider combination.
B) In order for there to be no net torque about the point where the wheel touches the road, the two forces acting at the center of mass, N and Ff, must have a ratio
Fr/N = tan theta = V^2/(gr)= 0.373
theta = 20.5 degrees
C) For the minimum turning radius at that velocity, Ff/N = µs = 0.65.
The leaning angle theta will be larger, and you will now have
V^2/(gr) = 0.65
r = V^2/(0.65 g)= 1.61 m
The above analysis makes an approximation that the centripetal acceleration is the same for all parts of the bike and rider;however for turning radii his small, the acceleration varies from top to bottom of the bike. Bike and rider are not small compared to the turning radius.
A) In order for the cyclist to remain balanced while making a turn, the net force acting on the bicycle and cyclist must point towards the center of the turn (centripetal force), which allows for circular motion. The normal force (N) is the force exerted by the road perpendicular to the tire's surface, and the friction force (Ffr) is the force exerted parallel to the tire's surface, opposing the motion. For the bicycle to remain balanced, the force of friction (Ffr) must provide the necessary centripetal force.
To explain why the angle theta must be given by tan(theta) = Ffr/N, we can consider the forces acting on the bicycle. The force of friction (Ffr) is the adjacent side and the normal force (N) is the opposite side of a right triangle formed by these forces. The tangent of an angle in a right triangle is defined as the ratio of the opposite side to the adjacent side. Therefore, we can say that tan(theta) = Ffr/N.
B) To find theta, we need to calculate the values of Ffr and N using the given information. We know that the centripetal force required to maintain circular motion is given by Fc = mv^2 / r, where m is the mass of the bicycle and cyclist. The centripetal force is provided by the force of friction (Ffr), so we can equate these two forces:
Ffr = mv^2 / r
Next, we need to calculate the normal force (N). The weight of the cyclist and bicycle is equal to mg, where g is the acceleration due to gravity. The vertical component of the weight is balanced by the normal force (N), so we can set up an equation:
mg = N * cos(theta)
Now we have two equations and two unknowns (Ffr and N), so we can solve them simultaneously. Substitute the expression for Ffr from the first equation into the second equation:
mg = (mv^2 / r) * cos(theta)
Now rearrange the equation to solve for theta:
theta = arccos((rg) / (v^2))
Plug in the given values of r = 2.8 m and v = 3.2 m/s to calculate theta.
C) To find the minimum turning radius, we need to use the equation for the maximum static friction force:
Ffrmax = µs * N
where µs is the coefficient of static friction between the tires and the road. We can substitute the expression for N from the previous equation into this equation:
Ffrmax = µs * (mg / cos(theta))
Since the maximum static friction force is equal to the centripetal force required for circular motion, we can equate equations:
mv^2 / r = µs * (mg / cos(theta))
Now, we need to solve this equation for the minimum turning radius (r). Rearrange the equation:
r = (µs * v^2) / (g * cos(theta))
Substitute the given values of µs = 0.65, v = 3.2 m/s, and theta from part B to calculate the minimum turning radius.