How Does Halving the Radius Affect the Period in Centripetal Acceleration?

AI Thread Summary
Halving the radius in a centripetal acceleration scenario affects the period significantly. If the acceleration remains constant, the new velocity must be calculated to maintain the same centripetal force. The initial and final equations for acceleration and period show that the period decreases, but the exact factor depends on the new velocity after the radius is halved. The discussion emphasizes the importance of maintaining constant acceleration to accurately determine the new period. Understanding these relationships is crucial for solving centripetal acceleration problems effectively.
smillphysics
Messages
28
Reaction score
0
What happens to the period when you cut the radius in half for a centripetal acceleration problem?


a=V^2/r
T=2pi*r/v


I need some background on this question. I believe the answer is the period is decreased by a factor of 2. I am just slightly confused.
 
Physics news on Phys.org
Well, that would depend. Are you keeping the acceleration on the body the same?
If so, then the problem isn't quite as trivial:
ainitial = v²/rinitial
afinal = u²/rfinal
Tinitial = 2πr/v
Tfinal = 2π(½r)/u

Since we want the acceleration before we cut the radius in half to be the same as after, all that remains to find u and Tfinal is a simple equation.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top