# General Rotation applied to a bicyclist in a turn

1. May 20, 2004

### Theelectricchild

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 am having some difficulty deriving why part a is true... and as for b I am thinking we should consider the "circular" translational motion of the bicycle and rider. Only problem is since I can't understand a, part b is somewhat confusing as well.

I would appreciate any help. Thanks a lot.

2. May 20, 2004

### arildno

"A bicyclist traveling with speed v = 3.2m/s on a flat toad"

3. May 20, 2004

### TALewis

Part a.

This is difficult without a free body diagram. Make sure you've drawn one.

Consider the head-on view of the bike and rider as a line at an angle to the horizontal and the vertical. I will assume that the view shows the bike leaning leftward. Let point A be where the tires meet the ground. Place the center of gravity G at some arbitrary point on the line. Let the distance AG be equal to d. Let theta be the angle formed between the bike line and a vertical line passing through A.

The weight mg acts downward through G. The normal force N acts upward at A. If the bike is leaning leftward, then point A wants to slip rightward. Therefore, the friction force Ffr acts leftward on A to oppose this impending motion.

To determine the relationship between theta, N, and Ffr, we will sum the moments about G, with a right-hand rule where the moment is positive if your thumb points out of the page. The sum of the moments must be zero if the bike has no tendancy to rotate in this plane (even though it is rotating in the horizontal plane):

\begin{align*} \sum M_G &= 0\\ 0 &= Nd\sin\theta - F_\textit{fr}\,d\cos\theta\\ Nd\sin\theta &= F_\textit{fr}\,d\cos\theta\\ \frac{F_\textit{fr}}{N} &= \tan\theta\\ \end{align}

4. May 20, 2004

### Divergent13

I have a question... Why was this thread moved? Since when does rotational dynamics fall outside of the mechanics aspect of physics?

5. May 20, 2004

### Staff: Mentor

It was moved here because it is more of a specific homework-type problem than a general discussion of rotational dynamics.

6. May 21, 2004

### Theelectricchild

I could use some more help on this one. I drew the FBD like you explained. I understand that M*Vy has to equal zero so the cyclist doesnt fall. I now have a triangle where Ffr is the side opposite to theta, N is the side adjacent, and side d is the hypotenuse. This gives me tan(theta)= Ff/N, sin(th)=Ff/d, and cos(th)= N/d. I can't see how you used these to obtain Ff*d*cos(th) etc.

7. May 21, 2004

### TALewis

I have made a figure that you can find here:

http://omega.uta.edu/~tal0701/bike.gif

What do you mean by M*Vy?

I don't think you can place d in the triangle with the two forces. For one thing, how can you write:

\begin{align*} \sin\theta&=F/d\\ \cos\theta&=N/d \end{array}

On the lefthand sides you have sines and cosines. Sines and cosines of angles are dimensionless (they have no units). On the righthand sides, you have forces divided by lengths. A newton divided by a meter is not dimensionless. In order for a physics equation to be valid, it must be "dimensionally homogeneous." That means the units on the left side must match the units on the right side.

Also, the term "moment" may not be used in your physics course. I remember in my introductory Physics I, only the term "torque" was used. They are essentially the same thing.

There might be other ways to solve this problem, but this is the only one that's fresh in my mind.

The moment of a force about a point is equal to the magnitude of the force times the perpendicular distance from the force's line of action to the point.

The sum of the moments about G must be equal to zero since there is no tendancy to rotate in this particular plane. The force mg has no moment about G since it acts through G.

The moment of N about G is equal to N*x, where x is the perpendicular distance between N and G. x is called the "moment arm." The moment of Ffr about G is equal to -Ffr*y. The negative sign comes from the fact that Ffr and N have a tendancy to rotate the bike in opposite directions: N would rotate the bike counterclockwise while Ffr would rotate it clockwise.

The moment equation again:

$$\sum M_G = 0 = Nx - F_\textit{fr}\,y$$

Also, from geometry:

\begin{align*} x&=d\sin\theta\\ y&=d\cos\theta \end{array}

You can plug these in to the moment equation and get the result that I had in the last post.

If this doesn't make sense, it would be helpful to know more about your physics course and what you have covered so far. It can be hard to answer a homework question at the appropriate level without knowing the context of the course.

8. May 21, 2004

### Theelectricchild

Ahhhh now it makes sense--- however I am wondering how to use this formula in solving theta...i assume the masses cancel in the equation since there is no mass given... I know choosing the CM makes the problem easier, but what must I set up? Thanks again for your post TA.

9. May 21, 2004

### TALewis

Do you know the formula for centripetal acceleration?

$$a_c=\frac{v^2}{r}$$

You know v (velocity) and r (radius).

What is the centripetal force that holds the bike in the circle? It is the friction force, Ffr. So you know:

$$F_\textit{fr}\, = ma_c$$

You also know that the normal force, N, equals the weight, mg:

$$N=mg$$

You determined that

$$\tan\theta=\frac{F_\textit{fr}}{N}$$

You should see that the mass will cancel out and you can solve for theta.

10. May 21, 2004

### Theelectricchild

Yeah right after my post i realized we set mv/r^2 and the masses will cancel since you equate the friction force to that. Thanks.