Is the Chain Adequate for the Amusement Park Ride?

[charan1]

Homework Statement

You have a new job designing rides for an amusement park. In one ride, the rider's chair is attached by a 9.0-m-long chain to the top of a tall rotating tower. The tower spins the chair and rider around at the rate of 1 rev every 4.0 s. In your design, you've assumed that the maximum possible combined weight of the chair and rider is 150 kg. You've found a great price for chain at the local discount store, but your supervisor wonders if the chain is strong enough. You contact the manufacturer and learn that the chain is rated to withstand a tension of 3000 N.

Will this chain be strong enough for the ride? Yes or no?

Homework Equations

F=(m)(w^2)(r)

The Attempt at a Solution

I converted 1rev/4seconds to rad/s-
.25rev/s x 2(pi) = 1.571rad/s

Then plugged it in and was confused!

F=150kg x 1.571 x r

what should I use for r?

or if i put 3000N in for F instead r comes to be 12.73m

I don't know what to do with these numbers someone please help!

Thank you

 

[Rake-MC]

scratch that

 

[Doc Al]

F=(m)(w^2)(r)

That describes the centripetal force.

What forces act on the chair+rider?

what should I use for r?

Express r in terms of the length of the chain and the angle it makes.

Hint: Write equations for vertical and horizontal forces.

 

[charan1]

Ok so I took a look at my FBD again and figured that

mg=cos(theta) x Max Tension

150g=cos(theta) x 3000

Theta= 60.7 degree's

and so r=cos(90-60.7) x 9m = 7.85m is my radius

then I plug it into the equation:

F=m x (w^2) x r

F=150kg x (1.571^2) x 7.85m

F=2,906.1N

2906N<3000N

So Yes Chain is strong enough

Is this correct?

 

[Doc Al]

Ok so I took a look at my FBD again and figured that

mg=cos(theta) x Max Tension

Instead, just figure out what the tension would actually be, then compare it to the max tension the chain can support. So this equation would be:
mg=cos(theta) x T

What's the other equation? (Since you don't know the angle, you'll need to combine both equations to solve for T.)

 

[charan1]

Yea i got the question wrong...but i still want to know how to do it properly.

was the way i got r questionable or something else?

What two equations was I suppose to combine?


I used
mg=cos(theta) x Max Tension

with 150kg in for m and 3000N in for Max Tension I solved for an angle which turned out to be 60.7 degrees since this is the angle between the y component of the tension and the tension vector I just subtracted it from 90 degree's to get the other side of the angle which is in turn the angle between the x component and the tension vector.

And so I used that angle in trig to get the x component of the right triangle made by the rope when it is swinging with the fact known that the rope is 9 meters long.

 

[Doc Al]

Yea i got the question wrong...but i still want to know how to do it properly.

was the way i got r questionable or something else?

I know what you did. I'm trying to explain what you should have done.

(Your method of finding "r" assumed that the tension was equal to the max value. That's not a valid assumption.)

You have all the data needed to find the tension in the chain for the given motion. Solve for that tension. (No need to make any assumptions or make use of the "max tension" value.) After you've found the tension, then you can compare it to that max value to see if this particular chain can be used.

What two equations was I suppose to combine?


I used
mg=cos(theta) x Max Tension

As I had said, correct that equation, which comes from analyzing the vertical forces on the chair/rider. Use:
mg=cos(theta) x T

The second equation will come from analyzing the horizontal force components. That's where you'll apply your knowledge of centripetal acceleration. And also a bit of trig to express the unknown "r" in terms of chain length L and angle.