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vorcil
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http://img13.imageshack.us/img13/9297/masteringphysicsq1.jpg

At the bottom it's 510N, Top 666N,

at bottom Fnormal=netforce= mv^2/r + mg

http://img13.imageshack.us/img13/8244/masteringphysicsq1g.jpg
-my attempt

I'm not sure how to figure out the acceleration
or determine the time for one loop on the graph.
 
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great 510 n = 5 seconds...
666n = 15 seconds
 
I don't think you need to determine the period or the velocity.

Where will the weight be the minimum? And the maximum?

Presuming that the ferris wheel is not a super spinning Whirl-a-Gig, then you know

mg - mv2/r = Min
mg + mv2/r = Max

Then just solve for m*g.
 
LowlyPion said:
I don't think you need to determine the period or the velocity.

Where will the weight be the minimum? And the maximum?

Presuming that the ferris wheel is not a super spinning Whirl-a-Gig, then you know

mg - mv2/r = Min
mg + mv2/r = Max

Then just solve for m*g.

I got two different awnsers, for the coaster at the bottom 12.171kg and top 14.52kg
i think I've done it wrong :\
 
vorcil said:
I got two different awnsers, for the coaster at the bottom 12.171kg and top 14.52kg
i think I've done it wrong :\

Try constructing the equations.

Then subtract 1 from the other.

You will determine then what mv2/r is and then you can figure the weight from either of the 2 equations.

I only get 1 answer.
 
LowlyPion said:
Try constructing the equations.

Then subtract 1 from the other.

You will determine then what mv2/r is and then you can figure the weight from either of the 2 equations.

I only get 1 answer.

What like?
666-510 = (mg + mv^2/r) - (mg - mv^2/r)
i can't figure out the velocity for the mv^2/r
 
666 = mg + mv^2/r
510 = mg - mv^2/r

666+510 = 2mg + - mv^2/4
= 1176 = 2mg
1176/9.8 = 120
120/2 = 60

60kg?
 
vorcil said:
666 = mg + mv^2/r
510 = mg - mv^2/r

666+510 = 2mg + - mv^2/4
= 1176 = 2mg
1176/9.8 = 120
120/2 = 60

60kg?

That's right. Adding them works too. In fact better as it yields the m*g directly.

m = 1176/(2*9.8) = 60