2
$\begingroup$

I have a general question regarding the centripetal force. In the example of a ferris wheel where there is a normal force pushing up against the person and the gravitational force pulling the person down, which force is centripetal? I know that the centripetal force counters the linear velocity, tangent to the circle of motion, which allows the object or person to stay in circular motion but which force is actually pulling it towards the center, the gravitational force or the normal force? Also, would it be correct to say that the net force equals zero (since the person is neither moving towards or away from the center) in this example or does the net force equal the centripetal force (since the centripetal force has to counter the linear velocity --- if this is correct, how would I compare the two since linear velocity is not a force)?

I know that if a car is moving around a banked curve, the horizontal normal force will be centripetal but what about in other examples such as the ferris wheel? Also would the net force of a car moving around a bank curved be zero since it is neither moving towards or away the center?

tl;dr - is the net force in a centripetal force example zero or is the net force equal to the centripetal force? Also, how would I relate this to the linear velocity that cancels it out?

Thanks for the help!

$\endgroup$
  • $\begingroup$ If a body is in uniform circular motion then the net force is radially inward.If the body move in a circle of fixed radius with variable speed then direction of net force is not towards the centre. $\endgroup$ – Paul Jan 11 '15 at 0:58
0
$\begingroup$

The component of the net force pointing to the center is the centripetal force. In your Ferris wheel example, we can choose three points: top (12 o'clock), bottom (6 o'clock) and halfway (3 o'clock). At 12, centripetal is normal force plus gravitational force; at bottom they are on opposite direction ($F_c=F_n-F_g$) and at 3 only the normal force points inwards while gravity speeds up or slows down the wheel'a rotation (depending on direction of motion).

$\endgroup$
  • $\begingroup$ If the centripetal force is a component of the net force, how do I find the centripetal force if I am just given the Net force? What else would I need to know? Thanks! $\endgroup$ – Ghost Koi Jan 11 '15 at 1:51
  • $\begingroup$ You need the direction of the net force (relative to the radial direction) to figure out the centripetal force $\endgroup$ – Floris Jan 11 '15 at 2:34
  • $\begingroup$ That makes sense, so if there is only one force acting on an object then that force is the centripetal force but if there are two forces acting on an object, the only centripetal force is the net force since the object would have to be moving uniformly in a circle. So, if I had the force of gravity and the normal force acting on an object in uniform circular motion, I would have to determine the value of both forces, add them together, and then be given the angle of the net force to be able to find the centripetal force. Is this correct? $\endgroup$ – Ghost Koi Jan 11 '15 at 3:58
  • $\begingroup$ Or, if I was not given the angle, I would just have to recognize that m(vˆ2/r) = mg + FN and then solve accordingly, correct? $\endgroup$ – Ghost Koi Jan 11 '15 at 4:05
  • $\begingroup$ Your equation (in your most recent comment) is only valid if you know the direction of the force! You could say $\frac{mv^2}{r}=F_n+mg\cos\theta$ where you can see $\theta=0$ at the top of the Ferris wheel, and $\theta=\pi$ at the bottom. $\endgroup$ – Floris Jan 11 '15 at 4:09
1
$\begingroup$

The centripetal force is not a physical force but rather the component of the force which points towards the center during circular motion. For the example of the Ferris wheel, the centripetal force depends on the position. For instance, if the the person is on the top of the ferris wheel, the gravitational and the normal force combined is the centripetal force, but if the person is in the bottom of the ferris wheel, the normal force minus the gravitational force is the centripetal force. The net force of a car moving around a bank curved is not zero, rather, because the net force is always pointing perpendicular to the velocity, the motion is circular and hence the car never moves toward or away from the center.

$\endgroup$
  • $\begingroup$ So, assuming the object is moving in circular motion, only the net force can be though of as centripetal? At the bottom of the ferris wheel, could I just subtract the two forces and then divide by the mass to find the radial acceleration? Essentially, I need to find all of the forces involved before I can calculate the centripetal force, is this correct? Thanks for the help! $\endgroup$ – Ghost Koi Jan 10 '15 at 23:50
  • $\begingroup$ Yes, you need to find all of the forces, but once you find the net force, only the component of the net force in the direction pointing towards the center is the centripetal force. The other component is the tangential force. $\endgroup$ – Bob Bobby Jan 10 '15 at 23:52
  • $\begingroup$ Okay. But how would I separate these two forces when calculating something like the radial acceleration. Also, Assuming I was looking at an object at the bottom of a Ferris Wheel and I subtracted the Normal force (mg) and the gravitational force ( also mg), wouldn't that just give me a net force of zero? What am I missing? Thanks again! $\endgroup$ – Ghost Koi Jan 10 '15 at 23:59
  • 1
    $\begingroup$ When at the bottom of the wheel, you "weigh" more. So the forces are not equal. The normal force is greater than gravity. That's why you accelerate up. You can calculate one from the other. If you have weight (normal force), you can calculate (radial) acceleration. If you have acceleration, you can calculate the normal force. $\endgroup$ – BowlOfRed Jan 11 '15 at 0:04
0
$\begingroup$

Centripetal Force when seen from a non inertial reference frame becomes centrifugal force. Now this is a pseudo force in this case and Pseudo forces do not belong to any category of forces. in an inertial frame it is just the net force that acts.... As in your case it would be gravity and normal reaction subtracted to give a net force that acts towards center.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.