Magnitude of avg acc. of train moving in a circle?

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  • #1
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Homework Statement


"Assume that a train is traveling at 65 cm/s, and the track has radius 20cm. What is the magnitude of the train's average acceleration as the train goes from the tree marked A to the next tree B, located exactly 1/4 of the way around the track from tree A? [Hint: This question is not asking for the train's centripetal acceleration.]"

Homework Equations


Equations of motion of object moving in circle, using ω,∝, θ.
aavg=(vf-vi)/Δt


The Attempt at a Solution


The answer is given to be 190 cm/s2 but why wouldn't it be zero? Maybe I'm misunderstanding the question, but if it doesn't want centripetal acceleration, it must mean tangential acceleration. And I thought that only the direction of tangential velocity changes, not magnitude. I also included a picture.
20160326_130018.jpg
 

Answers and Replies

  • #2
gneill
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Consider that velocity is a vector. Make a sketch showing the two velocity vectors of interest.
 
  • #3
34,685
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If the answer would be zero, the average acceleration would be zero, so initial and final velocity would be the same.

The answer is not the centripetal acceleration directly (and you don't have to calculate it), but this acceleration does contribute to the answer.
 
  • #5
gneill
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Would vi=<0,65> and vf=<65,0>?
Yes, those are two candidates for vi and vf (the particular vectors depend upon what you choose as the orientation of the track in your frame of reference). They will work for determining the average acceleration over the interval.
 
  • #6
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So if I subtract vi from vf, I get <65,-65>. To find time I used the formula acentripetal=v2/r=ω2r, and solved for ω by plugging in 65 cm/s as v and 20 cm as r and found ω to be 3.25rad/s. I converted this to rev/s and multiplied by 1/.25rev (since 1/4 the way around track) and got time to be 2.07s. Did I go wrong somewhere?
 
  • #7
gneill
Mentor
20,816
2,792
So if I subtract vi from vf, I get <65,-65>. To find time I used the formula acentripetal=v2/r=ω2r, and solved for ω by plugging in 65 cm/s as v and 20 cm as r and found ω to be 3.25rad/s. I converted this to rev/s and multiplied by 1/.25rev (since 1/4 the way around track) and got time to be 2.07s. Did I go wrong somewhere?
Your time is incorrect. You really went the long way around to try to calculate it.

Your angular velocity ω looks good. What is the angular distance of a quarter circle? Distance over velocity yields time...
 
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  • #8
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Ah, θ=90°=1.57rad. t=Δθ/ω. t=.48s. Then √(652+652) divided by .48s gives the 190cm/s2. Thank you!
 
  • #9
34,685
10,838
Note that you do not need ω. You can calculate the length of 1/4 of a circle with radius 20 cm, and divide that by the speed of the train. It gives the same result in a more direct way.
 

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