• tectactoe
In summary, the smallest radius of an unbanked track that a bicyclist can travel at a speed of 24 km/h with a coefficient of static friction of 0.30 is approximately 195 meters. The equation for this calculation is (us)(Fn) = m(v^2/r), where Fn is equal to the normal force, in this case equal to the downward pull of gravity on the bike/person. The forces in the vertical direction cancel, while the forces in the horizontal direction, on the line through the center of the track, cancel because friction from the ground balances the centripetal force. Friction is dependent on the normal force, but it acts horizontally.
tectactoe
"What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 24 km/h and the coefficient of static friction between tires and track is 0.30?"

v = 24 km/h
us = 0.30

I know

(us)(Fn) = m(v^2/r)

but I am stumped as to how to get r from this, so I am thinking maybe I have the wrong equation? I just am so unsure of what to do!

x_x

in this case what is the normal force equal to?

I think that's what I'm not sure about...

Wouldn't the normal force be equal to the downward pull of gravity on the bike/person in this case (mg)? In which case, mass is not known. =[

well in that case you got mass on both sides of the equation, and they are both the same mass right?

>_<

Wow. I didn't even realize that. I'm an idiot. Thank you kindly, sir.

Wait, somethings wrong... my answers aren't correct.

having

(us)(Fn) = m(v^2/r)

(us)(mg) = m(v^2/r)

canceling masses

(us)g = (v^2/r)

solving for r

r = (v^2) / (us)g

r = (24^2) / (.30)(9.8)

r = 576 / 2.94

r = 195

and that's not the right answer... =[

Does Fn = something other than just mg? I'm not sure, really!

Equilibrium of forces, F(net)=0 so F(friction)+F(centripetal)=0

I guess I'm just confused as to how that will help me find the radius of the track... I mean, the Forces in the y direction cancel, but how does that help with r? Yikes this problem seems so simple but it's been stumping me for hours.

The forces in the y (vertical) direction cancel because the ground provides the normal force that counteracts the gravitational force. The forces in the horizontal direction, on the line through the center of the track, cancel because friction from the ground balances the centripetal force. Friction is dependent on the normal force, but it acts horizontally.

As Mthees08 said, $F_{\text{friction}} = F_{\text{centripetal}} = \text{?}$. From the equation for centripetal force, think about what happens when radius decreases. What must also happen to the friction? How long can friction keep balancing centripetal force?

tectactoe said:
Wait, somethings wrong... my answers aren't correct.

having

(us)(Fn) = m(v^2/r)

(us)(mg) = m(v^2/r)

canceling masses

(us)g = (v^2/r)

solving for r

r = (v^2) / (us)g

r = (24^2) / (.30)(9.8)

r = 576 / 2.94

r = 195

and that's not the right answer... =[

Does Fn = something other than just mg? I'm not sure, really!

yep that's correct, but dotn forget that its: negative m*v^2/r, and negative Ff sice they both point inward toward the center

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## Question 1: What is centripetal force?

Centripetal force is a force that acts on an object moving in a circular path, directed towards the center of the circle. It is responsible for keeping the object in its circular motion.

## Question 2: How is centripetal force different from centrifugal force?

Centripetal force is the inward force that keeps an object moving in a circular path, while centrifugal force is the outward force that appears to push an object away from the center of rotation. In reality, centrifugal force is a fictitious force that only appears to exist due to the object's inertia.

## Question 3: How does friction affect centripetal force?

Friction can act as a centripetal force if it is directed towards the center of the circular motion. This can be seen in objects moving in a banked turn, where the friction between the object and the surface provides the necessary centripetal force to keep the object in its curved path.

## Question 4: What is the role of centripetal force in circular motion?

Centripetal force is essential for circular motion as it is responsible for keeping an object moving in a circular path and preventing it from moving in a straight line. Without centripetal force, an object would continue in a straight line tangent to the circle.

## Question 5: How does the magnitude of centripetal force change with velocity and radius?

The magnitude of centripetal force increases with an increase in velocity and a decrease in radius. This can be seen in the formula for centripetal force, F=mv²/r, where an increase in velocity (v) or a decrease in radius (r) will result in an increase in force (F).

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