centripetal acceleration/friction problem

[Genie133]

I need help with this problem:

What is the minimum radius that a cyclist can ride around without slipping at 10 km per hour if th coefficient of friction between hsi tires and the road is 0.5?

i got the wrong answer, and when trying to go back to figure it out, I don't know where to start!

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[super302]

1.58m

 

[BrokenGlass]

I need help with this problem:

What is the minimum radius that a cyclist can ride around without slipping at 10 km per hour if th coefficient of friction between hsi tires and the road is 0.5?

i got the wrong answer, and when trying to go back to figure it out, I don't know where to start!

F_s = m*V^2/r

N * mew_s = m*V^2/r

m * g * mew_s = m*V^2/r

g * mew_s = V^2/r

r = V^2 / (g * mew_s)

You do the math. I hate doing the math.

Since g in in SI units, convert km/hr to m/s before plugging these values into the equation. The answer will be in meters. If you want, convert the answer to km.

Note: centripetal force in this case is static friction.

 

[Genie133]

1.58m

according to the book the answer is 4.1

F_s = m*V^2/r

N * mew_s = m*V^2/r

m * g * mew_s = m*V^2/r

g * mew_s = V^2/r

r = V^2 / (g * mew_s)

You do the math. I hate doing the math.

Since g in in SI units, convert km/hr to m/s before plugging these values into the equation. The answer will be in meters. If you want, convert the answer to km.

Note: centripetal force in this case is static friction.

i think that's how i did it the first time, and i got the wrong answer. i plugged and chugged using your exact method, and got 1.48, but the answer in the back is 4.1m....

this looks to me to be the right way to do it, but it's not getting the right answer... what gives?

 

[Corpus Callosum]

according to the book the answer is 4.1




i think that's how i did it the first time, and i got the wrong answer. i plugged and chugged using your exact method, and got 1.48, but the answer in the back is 4.1m....

this looks to me to be the right way to do it, but it's not getting the right answer... what gives?

even if you were to consider the tangential acceleration and take the long route, eventually you get:


a^2= a^2 (tan) + a^2 (cent) where

a= 4.9 m/s^2 this came from F=ma

r= sqrt[v^4(1+16pi^2)/{16*4.9^2*pi^2}]

you plug in the numbers and the answer will still be: 1.58 meters. precisely why we don't have tangential acceleration, even if you were to consider it, it cancels out (it is zero in the first place for uniform circular motion) . I derived it just to confirm the validity of the answer others got and it has to be that.
Chances are the book made a mistake.

 

[changarang]

i agree. i got 1.54m without rounding (did all the calculator stuff at the end). i solved it the way BrokenGlass solved it.

 

[smp017]

why would centripical force equal static friction in this problem?
Is it because it keeps the person going in a circle?


I think i had a question like this when I took the MCAT on july 24 btw.

 

[Chowdder]

why would centripical force equal static friction in this problem?
Is it because it keeps the person going in a circle?


I think i had a question like this when I took the MCAT on july 24 btw.

Short answer, Yes.

 

[FIREitUP]

why would centripical force equal static friction in this problem?
Is it because it keeps the person going in a circle?


I think i had a question like this when I took the MCAT on july 24 btw.

because their are two forces acting upon this person, one being the centripetal force acting towards the center of the pathway, and the other being the friction force which points opposite. Since there is no slippage, the person is in equilibrium, so we say mv^2/r=staticfriction x normal force.

 

[pyromatic]

because their are two forces acting upon this person, one being the centripetal force acting towards the center of the pathway, and the other being the friction force which points opposite. Since there is no slippage, the person is in equilibrium, so we say mv^2/r=staticfriction x normal force.

This is incorrect. The person is clearly not in translation equlibrium as his velocity is changing (the magnitude might not be changing, but clearly the direction is), hence there must be a net force. There is only one force acting on the person and that is static friction, which is pushing him inward, along the radius of the path. Because he isn't sliding, it must be static friction.

Now think about a stationary mass, sitting on a table, and attached by a string to a device that measures force. If you pull horizontal, along the table, on the instrument, you will note that the force increases from 0 to some maximum (maximum of static friction) before the mass begins to move and then drops to some constant force while you move the mass at a constant speed (kinetic friction).

So static friction can be anything between 0 and its maximum determined the coefficient * the normal force. So you could turn your bike as slowly as you want without slipping as the force needed to pull you inward and "turn" you, provided by static friction, is less than its maximum. The fastest you can turn is determined by the maximum amount of acceleration the inward force can produce which is simply the maximum of static friction.

If there were an outward force, you'd need a stronger inward force to result in the same net force.

 

[FIREitUP]

This is incorrect. The person is clearly not in translation equlibrium as his velocity is changing (the magnitude might not be changing, but clearly the direction is), hence there must be a net force. There is only one force acting on the person and that is static friction, which is pushing him inward, along the radius of the path. Because he isn't sliding, it must be static friction.

Now think about a stationary mass, sitting on a table, and attached by a string to a device that measures force. If you pull horizontal, along the table, on the instrument, you will note that the force increases from 0 to some maximum (maximum of static friction) before the mass begins to move and then drops to some constant force while you move the mass at a constant speed (kinetic friction).

So static friction can be anything between 0 and its maximum determined the coefficient * the normal force. So you could turn your bike as slowly as you want without slipping as the force needed to pull you inward and "turn" you, provided by static friction, is less than its maximum. The fastest you can turn is determined by the maximum amount of acceleration the inward force can produce which is simply the maximum of static friction.

If there were an outward force, you'd need a stronger inward force to result in the same net force.

the fact that he is changing direction is taken to account by the centripetal force, and since the centripetal force is ALWAYS opposite the friction force, and that the bike is not slipping towards the middle (where the centripetal force is pointing), we can set them equal.

 

[pyromatic]

The centripetal force is the static friction. The centrifugal force is probably what you mean here. That is a pseudoforce that is introduced if you pretend the reference frame of the cyclist is inertial, which clearly it is not.

 

[FIREitUP]

the centripetal force may be equal in magnitude to the friction force, but it is opposite in direction. the centripetal force is the force that is inherent when you are traveling in a uniform circle (whether you're spinning a ball on a string or driving on a track).

i wish i could just draw a free body diagram, and then you would see that there are only two net forces acting upon the bike: the force of friction (which is perpendicular to the tangential velocity) and the centripetal force (which is also perpendicular to the tangential velocity, but is opposite the force of friction), and thus, since the bike is not slipping, they are equal to each other in MAGNITUDE but are opposite in direction.

okay, i dont really care, we're basically in agreement on how to solve the problem, so thats all that matters.

 

[pyromatic]

For your ball on a string example, the centripetal force is the tension in the string. Centripetal force is not "inherent;" it is what causes the circular motion. - the external force you must supply for a given radius and linear velocity in order for circular motion to result. It could be gravitational, electromagnetic, etc. I'm not exactly sure how you're under the impression the force of friction is pointing outward, but it isn't. If we have two REAL forces, that are equal and opposite, there is no net force, thus no acceleration and no circular motion.

 

[Phoenix 47]

For your ball on a string example, the centripetal force is the tension in the string. Centripetal force is not "inherent;" it is what causes the circular motion. - the external force you must supply for a given radius and linear velocity in order for circular motion to result. It could be gravitational, electromagnetic, etc. I'm not exactly sure how you're under the impression the force of friction is pointing outward, but it isn't. If we have two REAL forces, that are equal and opposite, there is no net force, thus no acceleration and no circular motion.

correct. centripetal force isnt some invisible hand, it has to be caused by something - in this case, static friction.

 

[thechairman]

is the static friction applying a torque to the wheel? is that what's causing the rotation (and hence, the centripetal accelaration)?

 

[FIREitUP]

For your ball on a string example, the centripetal force is the tension in the string. Centripetal force is not "inherent;" it is what causes the circular motion. - the external force you must supply for a given radius and linear velocity in order for circular motion to result. It could be gravitational, electromagnetic, etc. I'm not exactly sure how you're under the impression the force of friction is pointing outward, but it isn't. If we have two REAL forces, that are equal and opposite, there is no net force, thus no acceleration and no circular motion.

i understand. my bad.

 

[changarang]

for the people who asked about static friction...just be careful. if the question had asked about a car going around in a circle (no skidding), that's static friction that's causing the car to go around in a circle (the centripetal force is equal to the force of static friction). but if the question had asked about a car that slams on the brakes and then skids to a halt, that's kinetic friction because the tires are moving/sliding/slipping against the ground.