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:
 
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