Centripetal Force Loop-the-loop

[JoshBuntu]

Homework Statement

There is a loop-the-loop, thing, and a dude on a bicycle is going to ride around the loop. Total mass is 1 kg (I'm aware that isn't realistic at all...) and radius is 10 meters.

a) what is the minimum speed that the cyclist must have to make it over the top of the loop without falling off?
b) If this block has this minimum speed, then compute the apparent weight of the cyclist at the bottom of the loop.
c) what is the normal force on the cyclist at the 3 oclock position?

Homework Equations

a(centripetal)= (v^2)/r
F(gravity)=mg

The Attempt at a Solution

The answer I calculated to part a is 98.1 m/s. I assumed that Normal Force equals zero. For part b I got 972N . I assumed that F(normal) - F(gravity)=ma(centripetal)

I would really appreciate if someone could just double check those for me as I have no answer key.

I'm having significant problems with part c. I drew a free body diagram of the cyclist at the 3 oclock position. So normal force is acting to the left towards the center of the circle, with centripetal acceleration, and F(gravity) is acting downwards. If you're on a wall like that, then isn't normal force zero? But then I know that's not the case because the cycle is driving into the wall and normal force is pushing outward and...ok, so I don't really understand what the normal force is doing here. Could someone please explain the concept? I think if I understand the concept I could do the math, but I just don't understand it. Thanks!

 

[PhanthomJay]

Homework Statement

There is a loop-the-loop, thing, and a dude on a bicycle is going to ride around the loop. Total mass is 1 kg (I'm aware that isn't realistic at all...) and radius is 10 meters.

a) what is the minimum speed that the cyclist must have to make it over the top of the loop without falling off?
b) If this block has this minimum speed, then compute the apparent weight of the cyclist at the bottom of the loop.
c) what is the normal force on the cyclist at the 3 oclock position?

Homework Equations

a(centripetal)= (v^2)/r
F(gravity)=mg

The Attempt at a Solution

The answer I calculated to part a is 98.1 m/s. I assumed that Normal Force equals zero.

that's the correct assumption, but your answer is incorrect. Please show your work

For part b I got 972N . I assumed that F(normal) - F(gravity)=ma(centripetal)

again right assumption, wrong answer. Are you using conservation of energy to find the speed ? of course, you need to correct part a first.

I'm having significant problems with part c. I drew a free body diagram of the cyclist at the 3 oclock position. So normal force is acting to the left towards the center of the circle, with centripetal acceleration, and F(gravity) is acting downwards. If you're on a wall like that, then isn't normal force zero? But then I know that's not the case because the cycle is driving into the wall and normal force is pushing outward and...ok, so I don't really understand what the normal force is doing here. Could someone please explain the concept? I think if I understand the concept I could do the math, but I just don't understand it. Thanks!

The normal force acts perpendicular to the object (cyclist) at a contact point, so that is the force acting to the left due to the centripetal acceleration at that point. You again should use conservation of energy to find the speed at that point.

 

[JoshBuntu]

Conservation of energy? Ohh uhhh...is there another way? We didn't learn energy yet so I'm assuming there's another way to do this...

 

[PhanthomJay]

Apparently, then, the speed of the cyclist is assumed constant throughout the loop, so after you calculate the min speed at the top, use that same value of speed everywhere, using Newton 2 to calculate the normal force at the bottom and 3 o'clock position.

 

[rwisz]

Hello! As a fellow scientist, I can help clarify the concept of normal force and how it relates to the loop-the-loop scenario.

First, let's define normal force. It is the perpendicular force that a surface exerts on an object in contact with it. In this case, the surface is the track of the loop-the-loop and the object is the cyclist. Normal force is always perpendicular to the surface, so in this scenario, it would be pointing towards the center of the loop.

Now, let's look at the 3 o'clock position on the loop. At this point, the cyclist is moving horizontally and there is no centripetal acceleration, so the only force acting on the cyclist is gravity. This means that the normal force is equal in magnitude to the force of gravity, but in the opposite direction. The normal force is what keeps the cyclist from falling through the loop and provides the necessary centripetal force to keep the cyclist moving in a circular path.

To calculate the normal force at the 3 o'clock position, we can use the equation F(normal) = F(gravity) + ma(centripetal). Since there is no centripetal acceleration at this point, the equation simplifies to F(normal) = F(gravity). Plugging in the values, we get F(normal) = (1 kg)(9.8 m/s^2) = 9.8 N.

I hope this helps clarify the concept of normal force and how it applies to the loop-the-loop scenario. Keep up the good work in your studies!