Roller Coaster and Centripetal Acceleration

[hrithikguy]

Homework Statement

Part of a roller-coaster ride involves coasting down an incline and entering a loop 8.00 m in diameter. For safety considerations, the roller coaster speed at the top of the loop must be such that the force of the seat on a rider is equal in magnitude to the rider's weigt. From what height above the bottom of the loop must the roller coaster descend to satisfy this requirement?

Homework Equations

a_c = v^2/r
Conservation of Energy

The Attempt at a Solution

m* a_c = mg

m* v^2/r = mg
v^2/r = g
v^2 = rg

mgh = 1/2 mv^2 + mg * 8
9.8 * h = 1/2 * rg + 8 * 9.8

9.8 * h = 1/2 * 4 * 9.8 + 8 * 9.8
h = 2 + 8
h = 10

However, the answer is 12.

Please guide me to the solution.

Thanks!

 

[ehild]

The Attempt at a Solution

m* a_c = mg

m* v^2/r = mg

There act two forces on the rider, one is gravity (mg) and the other is the normal force N from the seat. The problem says that the seat must exert a force on the rider equal to mg. The resultant force is equal to the centripetal force.


ehild

 

[hrithikguy]

So does this mean that n = -mg, so

$m * a_c = 2mg ?$

$a_c = 2g$$a_c = 19.6$

$v^2/r = 19.6$
$v^2 = 19.6 * 4 = 78.4$
$v = 8.85$

$9.8 * h = 1/2 * 9.85^2 + 8 * 9.8$

$x = 12.95$

Hmm.. this is closer, but still not correct.

 

[ehild]

You used v = 9.85 instead of 8.85. It is much better if you do not evaluate the equations, but express the height h in terms of R.

ehild

 

[hrithikguy]

Ah thank you! I plugged in 8.85 now and it came out to 11.996, which is close enough.
Thanks!

 

[ehild]

I show the other way: mv^2/R=2mg
hmg=mg*2R+1/2 mv^2 -->

v^2=2gR,
hg=2gR+1/2v^2--->hg = 3gR ---> h=3R.

ehild