# Aparent weight

1. Mar 8, 2005

### sevens

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 aparent 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.

2. Mar 8, 2005

### clive

Try this:
$$G_a=mg-F_{cf}$$
where
$$F_{cf}=\frac{mv^2}{R}$$
and $$v=2v_{critical}$$

The ratio is given then by $$\frac{G_a}{mg}$$.

Last edited: Mar 8, 2005
3. Mar 25, 2005

### Aim

I did the same thing you did, and the answer is 1/4.

It's still not right

4. Mar 25, 2005

### Aim

What is $$G_a$$ ?

5. Mar 26, 2005

### Staff: Mentor

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:
$F_n = mv^2/r - w$

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.