Cyclist cycling in a perfect circle - find radius

In summary: F} Horizontal Component ("F" cos30) balances "m*g" so rider is in equilibrium. {F} Vertical Component ("F" sin30) provides centripetal force for uniform circular motion.In summary, a 78 kilogram cyclist with a velocity of 12m/sec and leaning at 30 degrees in a perfect equilibrum is cycling in a perfect circle with a radius of 25.4 meters. This is found by using the formula F=mv2/r and solving for r, with the requirement that the rider is in equilibrium and the road surface provides both a horizontal component to balance the rider's gravitational force and a vertical component to provide centripetal force.
  • #1
Woody11
3
0
I am doing a Biomechanics course and it is just like physics. I don't have a large back ground when it comes to physics. So I need some help. I am having a problem with a question on an assignment. This is the question:

A 78 kilogram cyclist is cycling in a perfect circle, with a velocity of 12m/sec. If The cyclist is leaning at 30 degrees and is in a perfect equilibrum, what is the radius of teh circle they are cycling around?

It doesn't seem to be a hard question, but I have been trying to figure this out for about 5 hours now and nothing is hitting me on how to do it. I know it has to deal with centripetal force and I think I should be using the formula F=mv2/r at some point. But I have no idea what the first step is. I hope someone can help me get started, remember I have very little background when it comes to physics it has been a while since I have done it.

Thanks,
Woody :frown:
 
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  • #2
A 78 kilogram cyclist is cycling in a perfect circle, with a velocity of 12m/sec. If The cyclist is leaning at 30 degrees and is in a perfect equilibrum, what is the radius of teh circle they are cycling around?
Because rider is in equilibrium at 30 deg to the VERTICAL, the road surface applies Force "F" (shown in drawing below) whose components are sufficient to balance rider's gravitational force (m*g) and provide centripetal force (m*v2/r) required for uniform circular motion. Force "F" components are shown in the drawing. Thus, for equilibrium:
{Centripetal Force} = {"F" Horizontal Component}
::: ⇒ m*v2/r = m*g*tan(30)
::: ⇒ r = v2/{g*tan(30)}
::: ⇒ r = (12 m/sec)2/{(9.81 m/sec2)*tan(30)}
::: ⇒ r = (25.4 meters)

Code:
                       ^         ^
                       |        /  
                       |       / [COLOR=Red][B]F[/B][/COLOR]
                       |  30  /
                       |     /
                   m*g |    / m*g/cos(30)
                       |   /
                       |  /
                       | /
                       |/________>
 
                        m*g*tan(30)               


            [COLOR=Blue]Road Surface Force Components On Rider[/COLOR]
 
Last edited:
  • #3


Hello Woody,

First of all, don't worry if you don't have a strong background in physics. Biomechanics and physics go hand in hand, and it's perfectly normal to need some help with understanding concepts and solving problems.

For this question, you are correct in thinking that you will need to use the formula F=mv^2/r. This is known as the centripetal force formula and is used to calculate the force required to keep an object moving in a circular motion.

The first step is to identify the known values in the problem. In this case, we know the mass of the cyclist (m=78kg), the velocity (v=12m/sec), and the angle at which the cyclist is leaning (30 degrees). We also know that the cyclist is in perfect equilibrium, meaning that the forces acting on them are balanced.

Next, we need to draw a free body diagram to visualize the forces acting on the cyclist. In this case, there are two forces acting on the cyclist: the centripetal force (F) and the force of gravity (mg). The force of gravity is acting downwards, while the centripetal force is acting towards the center of the circle.

Now, we can use the formula F=mv^2/r to solve for the radius. Since the cyclist is in equilibrium, the centripetal force must be equal to the force of gravity. So, we can set up the following equation:

F=mg=mv^2/r

We can rearrange this equation to solve for r:

r=mv^2/mg

Plugging in the known values, we get:

r=(78kg)(12m/sec)^2/(78kg)(9.8m/s^2)

Simplifying, we get:

r=14.4m/s^2/9.8m/s^2

r=1.47m

So, the radius of the circle that the cyclist is cycling around is approximately 1.47 meters.

I hope this helps you understand the problem and how to approach it. Remember, it's always helpful to draw diagrams and identify known values before solving a physics problem. Good luck with your assignment!
 

1. What is the formula for finding the radius of a perfect circle?

The formula for finding the radius of a perfect circle is r = c/2π, where r is the radius, c is the circumference, and π is approximately 3.14.

2. How do you measure the circumference of a perfect circle?

The circumference of a perfect circle can be measured using a measuring tape or ruler. Simply wrap the measuring tool around the circle, making sure it is taut, and record the measurement in inches or centimeters.

3. Can the radius of a perfect circle be negative?

No, the radius of a perfect circle cannot be negative. It represents the distance from the center of the circle to any point on the circle's edge, and distance cannot be negative.

4. Why is it important to find the radius of a perfect circle in cycling?

In cycling, the radius of a perfect circle is important because it determines the trajectory and speed of the cyclist. A larger radius means a larger circle, which requires the cyclist to travel a longer distance at a slower speed. A smaller radius means a smaller circle, which allows the cyclist to travel a shorter distance at a faster speed.

5. How does the radius of a perfect circle affect the cyclist's balance?

The radius of a perfect circle does not directly affect the cyclist's balance. However, a larger radius may require the cyclist to lean more to maintain balance while turning, whereas a smaller radius may require less leaning. The cyclist's balance is primarily influenced by their speed and center of gravity.

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