Normal Force during centripetal acceleration

AI Thread Summary
When swinging a ball on a string, the existence of a normal force depends on the observer's perspective. An outside observer sees no normal force, as tension in the string accounts for the centripetal motion. However, from the ball's perspective, a pseudoforce appears to act outward, but this does not constitute a normal force since it lacks contact with a surface. The discussion emphasizes that normal forces require a physical surface for contact, and the concept of pseudoforces is relevant in non-inertial reference frames. Understanding these principles clarifies the dynamics of centripetal acceleration in different frames of reference.
prospectus
Messages
2
Reaction score
0
General Question Here:

Let's say you are swinging a ball around on a string. Will there be a normal force?
 
Physics news on Phys.org
That's a good question and it depends on your perspective.

If you take the view point of an outside observer, then no. There is the tension in the string, the circular speed and the radial acceleration; all motion is accounted for from that perspective.

If your view point is from the ball, then yes. There is a pseudoforce acting on you that pulls you away from the center. That pseudoforce is the normal force, which is opposite from the acceleration.

It's not a topic that I have much depth on, but that is what I know.

Research non-inertial reference frames for more.
 
Thank you. It makes sense that there would be no normal force because there is no surface acting on the ball.
 
You're welcome.

I'd just like to append what I said with this:
What I described from the view point of the ball would not be a normal force either. That is still a pseudoforce.
A normal force always requires contact with a surface.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top