Solving Rotations Homework: Double Speed vs Halving Radius

  • Thread starter Thread starter baobao11
  • Start date Start date
  • Tags Tags
    Rotations
AI Thread Summary
The discussion focuses on the effects of doubling speed versus halving the radius on a truck's acceleration around a circular track. Using the equation a = v^2/R, calculations show that doubling the speed from 15 m/s to 30 m/s results in an acceleration of 9 m/s^2, while halving the radius from 100 m to 50 m yields an acceleration of 4.5 m/s^2. The consensus is that doubling the speed has a greater impact on acceleration than reducing the radius. This is confirmed by the principle that any factor multiplied to velocity affects acceleration quadratically. Overall, the conclusion is that increasing speed significantly enhances acceleration more than decreasing radius.
baobao11
Messages
2
Reaction score
0

Homework Statement


A truck was driven around a circular track. Which would have a greater effect on the magnitude of its acceleration: moving to a track with half the radius or doubling the speed.
Show proof/sample calculations.

Homework Equations


I'm not quite sure but I went ahead and tried this equation:
a = v^2/R

The Attempt at a Solution


I tried plugging in numbers:
a = v^2/R
v= 15m/s
R = 100m
a= (15^2)/(100) = 2.25 m/s^2

doubled speed:
a= (30^2)/100 = 9 m/s^2

half radius
a = (15^2)/50 = 4.5 m/s^2

If I'm correct, doubling speed will affect the magnitude of the acceleration more than the halved radius, right?
Thank you so much!
 
Physics news on Phys.org
Yes that is correct since whatever factor you multiply the velocity by, becomes squared as well.

a=v2/r

if 'v' is replaced by 'kv', then a' = k2 (v2/r) = k2a
 
rock.freak667 said:
Yes that is correct since whatever factor you multiply the velocity by, becomes squared as well.

a=v2/r

if 'v' is replaced by 'kv', then a' = k2 (v2/r) = k2a


thank you very much for confirming my answer! :smile:
 
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 '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