# Centripedal acceleration question

1. Dec 31, 2005

### dragon513

Q. A BMX rider is about to ride over a large bump of radius R. What is the maximum speed the rider can travel without leaving the ground?

Choice
A. g/R
B. gR
C. (gR)^0.5
D. It depends on the mass of the rider

The answer provided is C, but howcome g, gravity, is the centripedal acceleration?

I was able to obtain the answer C by using this (V^2)/R = g = centripedal accel (C.A.).

However, shouldn't the direction of the C.A. be perpendicular to the direction the rider is moving to?

g is in fact the C.A. at the top of the large bump, but when the rider is going up or coming down, I don't think g is the C.A.

Am I not correct?

Thank you and Happy New Year!

2. Dec 31, 2005

### scholzie

Well, the centripital FORCE must equal the normal force of the rider, or he'll go flying off the bump, right? We care about the MAXIMUM velocity, which means we have to look at the case of highest normal force. Well that force is at a maximum when the biker is on the very top of the bump (when all of the force of the rider is normal to the bump). So, at that point, $F_{normal}=m_{rider}g$, and that must equal centripital force.
$$m_{rider}g=m_{rider}\frac{v^2}{r}$$

The masses cancel as they always do in cases similar to this, and simple algebra yields $v=\sqrt{gr}$

Make sense?

3. Dec 31, 2005

### dragon513

I got it now! Thank you!

4. Jan 1, 2006