Centripetal and average acceleration

AI Thread Summary
Centripetal acceleration is greater for Runner A on the inside lane due to the sharper turn required at the same speed compared to Runner B on the outside lane. While both runners experience the same change in velocity over a complete lap, Runner A achieves this in less time, resulting in a larger instantaneous centripetal acceleration. Average acceleration, calculated as the change in velocity divided by time, is zero for both runners after one full lap since their velocities return to the starting point. Therefore, while Runner A has a higher centripetal acceleration, both runners have the same average acceleration over the course of the lap. Understanding these concepts clarifies the dynamics of circular motion for objects traveling at constant speed.
ViewtifulBeau
Messages
50
Reaction score
0
This question is all qualitative, but I don't know what centripetal and average acceleration is around a circle. The question is about two runners on a circular track, Runner A on the inside lane, Runner B on the outside lane. The runners have the same SPEED, I think that runner A has a larger centripetal and average acceleration because A has to make a sharper turn. But I think that A and B undergo the same change in velocity as the go around the whole track, even though A will reach the end first. Are my assumptions correct? Thanks
 
Physics news on Phys.org
You assumptions are correct, yes.

runner A will have a greater accln because that runner is on a smaller radius, which, in effect, is the same as saying that he makes a sharper turn.

Average accln is the change in velocity divided by the time taken. They both have the same change in velocity, but runner B has taken longer, therefore has a smaller average accln.

If the measurements are taken over one lap though, then they both have the same average accln - it equals zero!
 
ViewtifulBeau said:
This question is all qualitative, but I don't know what centripetal and average acceleration is around a circle. The question is about two runners on a circular track, Runner A on the inside lane, Runner B on the outside lane. The runners have the same SPEED, I think that runner A has a larger centripetal and average acceleration because A has to make a sharper turn. But I think that A and B undergo the same change in velocity as the go around the whole track, even though A will reach the end first. Are my assumptions correct? Thanks

If two bodies are traveling around a common centre point and one is 'on the outside track' in relation to the other and they are both traveling at the same speed, the instantaneous centripetal acceleration will be greater for the body on the 'inside track' than the body on the 'outside track'. This can be easily seen because the outer body will lag behind the inner body if each have the same linear speed and so the inner body will have an equal change in velocity in less time than the outer body (or will have a larger change in velocity in the same time).
After one full revolution each, the average acceleration is zero for each body.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging 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