Speed of a bicycle - circles

[sauri]

Suppose you ride a bicycle on a horizontal surface in a circle with a radius of 20 m. The resultant force exerted by the road on the bicycle (normal force plus frictional force) makes an angle of 15 degs with the vertical.
a. What is your speed?
b. If the frictional force is half its maximum value, what is the coefficient of static friction?

The problem is I got 2 equations that i can think of to solve for a). One equation is a=V^2/r and the other is F=mrw^2. But for the second equation I don't know the w(angular vel) and for the first I don't know a(acceleration). So how do I solve this?

 

[tony873004]

Your rotational force is
$$\frac{Mv^2}{r}$$
Can you use this to find an expression for your frictional force? (hint: no math needed)

 

[kyle8921]

I would approach this problem by first understanding the physical principles involved. In this case, we are dealing with circular motion and the forces acting on the bicycle. From this, we can use the equations of motion and Newton's laws to solve for the speed and coefficient of static friction.

a. To find the speed, we can use the equation v^2 = rω^2, where v is the linear speed, r is the radius of the circle, and ω is the angular velocity. We can also use the equation F = ma, where F is the resultant force, m is the mass of the bicycle, and a is the centripetal acceleration. Since we know the angle between the resultant force and the vertical, we can use trigonometry to find the horizontal component of the force, which is equal to ma. This can then be equated to the frictional force, which is equal to μN, where μ is the coefficient of static friction and N is the normal force. Solving for v, we get v = √(μgrtanθ), where g is the acceleration due to gravity and θ is the angle between the resultant force and the vertical. Plugging in the values given, we get v = √(μ(9.8 m/s^2)(20 m)tan(15°)) = 5.3 m/s.

b. To find the coefficient of static friction, we can use the same equation as in part a, but this time we know that the frictional force is half its maximum value, which is equal to μmaxN. So, we can write the equation as F = μmaxN/2 = μN. Solving for μ, we get μ = F/N = (mg/2)/(mgcosθ) = tanθ/2. Plugging in the values given, we get μ = tan(15°)/2 = 0.13.

In summary, to solve for the speed and coefficient of static friction, we used the equations of motion, Newton's laws, and trigonometry to relate the forces and acceleration to the variables we are looking for. This approach can be applied to various problems involving circular motion and forces.