Roller Coaster Car: Apparent Weight vs. True Weight

AI Thread Summary
The discussion centers on calculating the ratio of a roller coaster car's apparent weight to its true weight when traveling at twice the critical speed. Initial calculations incorrectly suggested a 4 to 1 ratio, but further analysis revealed that the apparent weight is derived from the centripetal force and the car's true weight. The correct approach involves applying Newton's second law and recognizing that at critical speed, the apparent weight is zero. When the car moves at twice the critical speed, the apparent weight is calculated using the formula for centripetal acceleration. Ultimately, the correct ratio of apparent weight to true weight is determined to be 1/4.
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A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed.

What is the ratio of the car's apparent weight to its true weight?


from my notes i have that Velocity_critical= squareroot ( r w / m)

v = sqrt(rw/m)

so what i did was multiply v times 2 because the problem states were going at twice the critical speed.


so now my equation looked like

2v = sqrt(rw/m)

I squared both sides

4v^2 = rw/m

and solved for weight.

w=(4mv^2)/r

so i thouth that the apparent weight would be 4 times the actual weight, giving me a 4 to 1 ratio. this seems to be wrong. anybody know were i took a wrong turn. any help would be apreciated.
thanks.
 
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Try this:
G_a=mg-F_{cf}
where
F_{cf}=\frac{mv^2}{R}
and v=2v_{critical}

The ratio is given then by \frac{G_a}{mg}.
 
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I did the same thing you did, and the answer is 1/4.

It's still not right :confused:
 
clive said:
Try this:
G_a=mg-F_{cf}
where
F_{cf}=\frac{mv^2}{R}
and v=2v_{critical}

The ratio is given then by \frac{G_a}{mg}.
What is G_a ? :confused: :confused:
 
sevens said:
from my notes i have that Velocity_critical= squareroot ( r w / m)

v = sqrt(rw/m)
In this formula, w stands for the actual weight (w = mg), not the apparent weight. The apparent weight equals the normal force that the track exerts on the the car. (More accurately, the apparent weight is the reaction force to the normal force.) By definition of "critical speed", if the car is moving at the critical speed then the apparent weight at the top of the motion is zero, since there is no normal force.

Similar to what clive explained, consider the forces acting on the car:
- normal force, acting down (this is the apparent weight)
- real weight, acting down (w = mg)

Now apply Newton's 2nd law with centripetal acceleration:
F_n + w = mv^2/r (I take down as positive.)

So the apparent weight is:
F_n = mv^2/r - w

so what i did was multiply v times 2 because the problem states were going at twice the critical speed.


so now my equation looked like

2v = sqrt(rw/m)
No. If the critical speed is sqrt(rw/m), then twice the critical speed is 2sqrt(rw/m). Now use that value in the correct formula for apparent weight given above.
 
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