Centripedal acceleration question

[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!

 

[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?

 

[dragon513]

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?

I got it now! Thank you!

 

[Chi Meson]

Right answer, wrong reason.

Centripetal force is not a type of force itself, but a description of the direction of a force. Some "real" force must supply the required centripetal force in order to change direction at the proper rate.

When undergoing perfect circular motion, the centripetal force is the only unbalanced force on the object going in a circle. Therefore the centripetal force is also the net force.

In the case of the motercylce going over a bump, the maximum speed that he can attain is that which causes the normal force at the top of the bump to just become zero. When the normal force is zero at the top of the bump, the total force on the guy is due to gravity alone. This is why the centripetal force is set equal to the guy's weight. Any faster, and gravity could not supply a force great enough to maintain the circular path of the bump, and the bike will leave the ground.