Centripital force - did i do it correctly?

[klm]

centripetal force - did i do it correctly?

In an amusement park ride called The Roundup, passengers stand inside a 19.0 m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in the figure View Figure .

Suppose the ring rotates once every 4.40 s. If a rider's mass is 59.0 kg, with how much force does the ring push on her at the top of the ride?

Suppose the ring rotates once every 4.40 s. If a rider's mass is 59.0 kg, with how much force does the ring push on her at the bottom of the ride?

okay so I am not sure if i did this correctly, but this is what I have done so far:

Fc= m(V^2) / r

and V=(2 x pi x r) / t

so i got v = 2pi8/ 4.4 = 11.4
and then
Fc= 59(11.4^2) / 8 = 958

is that correct? and what is the diff. b/w the force at the top and bottom

THANKS!

 

[Doc Al]

Start by identifying the forces acting on the person at the top of the ride and apply Newton's 2nd law. Draw yourself a diagram. The acceleration is centripetal, so you'll need to use that formula at some point.

Also: If the diameter is 19 m, what's the radius?

 

[klm]

oh crap its 9.5 not 8. sorry i don't know what i was thinking!

 

[klm]

aren't the only forces acting on them gravity(weight) and normal force that is pushing on them from the ring?

 

[Doc Al]

aren't the only forces acting on them gravity(weight) and normal force that is pushing on them from the ring?

Right! So set up the equation for net force.

 

[klm]

okay this might be incorrect, because I am kind of confused about how centripetal acceleration and all of this really works...

so basically all i have for my free body diagraphm is Nforce pointing straight up on the y-axis and Fg pointing straight down..would that be right?

so Fnety= N-Fg= m(ay) ... N-mg= m (ay) and we know m= 59 and g=9.8
but i don't know what to do now.

does that (ay) mean (ar)= V^2 / r ?
i hope that isn't too confusing!

 

[klm]

n - mg = m ( V^2)/ r

m=59
v = 2 pi r / t
so 2 pi 9.5 / 4.4 = 13.56
r= 9.5
g=9.8

so n = 1721.157 ??

 

[klm]

would that be the correct way to use the formula?

 

[Doc Al]

n - mg = m ( V^2)/ r

Almost: Which way does the normal force act at the top? (And what about at the bottom? That's what makes the difference.)

 

[klm]

umm i am not really sure. i know normal has to push back against the person. so like the ring is pushing back against the person. so maybe normal force goes horizontally?

 

[klm]

i don't understand how normal force is going to change from the top and bottom.

 

[Doc Al]

The normal force is always perpendicular to the surface. When you're at the top, where's the surface with respect to you? Which way must it be pushing?

 

[klm]

so is the ring considered to be the surface, so the ring is pushing horizontally. so in the positive x direction at the top?

 

[klm]

no wait, would it be in the neg x direction at top, and positive x direction at the bottom

 

[klm]

i have to go now, but i hope that maybe tomorrow you can help me some more and help me understand about this normal force!

 

[klm]

hi doc al, do you think you can help me finish this problem please!

 

[Doc Al]

no wait, would it be in the neg x direction at top, and positive x direction at the bottom

Exactly right.

 

[klm]

ahh thank you! i think i understand.
so does that mean i have to use Fnetx = -N = m ( V^2)/ r
and then Fnety = -mg = m ( V^2)/ r

but shouldn't one of the Forces = 0?

 

[Doc Al]

Not exactly.

At the top, the net force is: -N -mg
So Newton's 2nd law gives:
-N -mg = -mv^2/r (note that the acceleration points down so is also negative)

At the bottom, the net force is: +N -mg

 

[klm]

ooh. but can you maybe explain a little on why acc is neg. i know you wrote that it is pointing down, but isn't that already considered when you draw the free body diagram, so you wouldn't have to make it neg in the equation?

 

[Doc Al]

Whether something's negative or not just depends upon your sign convention. We are making everything that points down be negative: weight, normal force, and acceleration. The most important thing is to be consistent.

How would the signs change for the bottom position?

 

[klm]

N will be positive
Fg will remain negative
and Acc will be positive.. i think because you are moving up, so acc up..
so

Fnet= N-mg= m(v^2)/r at the bottom

 

[Doc Al]

Exactly! (The acceleration is positive because it points towards the center, which is up when at the bottom.)

 

[klm]

thank you Doc Al! i really appreciate the help!

 

[klm]

i have one more question for this problem that i don't even know how to begin , What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

i don't understand what the question is asking or asking me to solve?

 

[Doc Al]

Hint: When they start to fall off, what happens to the normal force?

 

[klm]

umm well there wouldn't be any Normal force right? b/c they arent touching the wheel anymore. or is that completely wrong...

 

[Doc Al]

You got it. Solve for the speed that makes the normal force at the top just equal zero. That will be the minimum speed to keep them from falling out.

 

[klm]

so would my equation just be -mg = -mv^2/r ?

 

[Doc Al]

so would my equation just be -mg = -mv^2/r ?

Exactly!

 

[klm]

yay! thank you so much!

 

[klm]

oh no , am i doing something wrong.

don't the m's cancel out on both sides. so you get g=V^2 / r
so 9.8= v^2 / 9.5
so v = 9.65

 

[Doc Al]

Good! Now use that to find the the corresponding period.

(The mass canceling just means that the answer does not depend on the mass.)

 

[klm]

i'm sorry i don't know how to do that? isn't a period just 2xpixr or am i thinking of the wrong thing

 

[Doc Al]

The period is the time it takes to go around one complete spin. You have the speed; the distance is one circumference.

 

[klm]

i'm sorry i still do not understand. so i need to multiply the speed (9.65) by the distance (2PiR) ?

 

[Doc Al]

You should know this: Distance = speed X time.

 

[klm]

ooh i see what you are saying. I'm sorry i should have realized that.