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Linear Acceleration at the Top and Bottom of a wheel

  • Thread starter Eggyu
  • Start date
  • #1
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hey guys, i was wondering if anyone could explain how to find the linear acceleration of a point at the top and bottom of a wheel. the homework problem is as follows (just so you know the context, i dont want the answers, i want to know how to go about this.):

Consider a 61 cm diameter tire on a car traveling at 50 km/h on a level road in the positive direction of an x axis.

(a) Relative to a woman in the car, what is the linear velocity v of the center of the wheel.
m/s
(b) What is the magnitude a of the linear acceleration?
m/s2
(c) What is v for a point at the top of the tire?
m/s
(d) What is a for a point at the top of the tire?
m/s2
(e) What is v for a point at the bottom of the tire?
m/s
(f)What is a for a point at the bottom of the tire?
m/s2


Now repeat the questions relative to a hitchhiker sitting near the road.

(g) What is v at the wheel's center?
m/s
(h) What is a at the wheel's center?
m/s2
(i) What is v at the tire top?
m/s
(j) What is a at the tire top?
m/s2
(k) What is v at the tire bottom?
m/s
(l) What is a at the tire bottom?
m/s2
 

Answers and Replies

  • #2
338
0
[tex]v=\omega r[/tex]
[tex]a_{cent}=\frac{v^2}{r}[/tex]

You should be able to figure out the rest.
 
  • #3
6
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Thanks for the help, it worked the the two on the first part but it didn't work for the second set of two.
 
  • #4
338
0
Thanks for the help, it worked the the two on the first part but it didn't work for the second set of two.
I think that should have worked for the hitchhiker.

For the woman, the relative linear velocity of the center of the wheel is 0 (it's going the same speed as she is).

The bottom of the wheel will be -v, and the top +v.
 
  • #5
6
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Yeah, i plugged the answers into the site and it did not accept them as correct. Within the hitch hiker part, the accel would be zero (according to the a = v^2/r) because the v is zero at the bottom. For the hitch hiker/top of the wheel part, i calculated the accel to be 2.53E3 m/s^2, which is obviously not right.
 
  • #6
338
0
Yeah, i plugged the answers into the site and it did not accept them as correct. Within the hitch hiker part, the accel would be zero (according to the a = v^2/r) because the v is zero at the bottom. For the hitch hiker/top of the wheel part, i calculated the accel to be 2.53E3 m/s^2, which is obviously not right.
Regardless of frame of reference among the hitchhiker or the woman, any point on the wheel, except the center, is going to be accelerating towards the center of the wheel.

That centripetal acceleration is given by:

[tex]a_{cent}=\frac{v^2}{r}=\omega^2 r[/tex]
 
  • #7
338
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Also make sure your units are correct.

You were given the diameter in cm, and the velocity in km/h.

Convert everything to meters, seconds and radians.
 
  • #8
6
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Yeah, i just rechecked my work and everything at it worked out. You freaking rock man! Thanks for all the help!
 
  • #9
338
0
http://www.turboconnection.com/pics/smilies/thankya.gif [Broken]
 
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