Another roller coaster question

In summary, the weight of the coaster at the bottom of the ferris wheel is 60kg, while at the top it is 60kg. To determine this, equations were constructed and solved, leading to the equation 666 = mg + mv^2/r, 510 = mg - mv^2/r, and a final weight of 60kg.
  • #1
vorcil
398
0
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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  • #2
great 510 n = 5 seconds...
666n = 15 seconds
 
  • #3
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.
 
  • #4
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 :\
 
  • #5
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.
 
  • #6
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
 
  • #7
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?
 
  • #8
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
 

What is the average speed of a roller coaster?

The average speed of a roller coaster can vary greatly depending on the specific ride, but it is usually between 25-40 miles per hour.

How do roller coasters stay on the track?

Roller coasters use a combination of gravity and centripetal force to stay on the track. The wheels on the coaster cars are also designed to hug the track tightly to prevent the cars from flying off.

What is the tallest roller coaster in the world?

As of 2021, the tallest roller coaster in the world is the Kingda Ka at Six Flags Great Adventure in New Jersey, USA. It stands at 456 feet tall.

How are roller coasters tested for safety?

Roller coasters undergo rigorous testing before they are opened to the public. This includes computer simulations, physical tests, and even test rides with dummies to ensure the safety of riders.

How do roller coasters loop without falling off the track?

Roller coasters use a combination of speed, momentum, and centripetal force to complete loops without falling off the track. The shape of the loop and the design of the wheels also play a crucial role in keeping the cars on the track.

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