Square wheel and the speed of the point

In summary, the conversation discusses a bumpy road with a specific shape for each hump, a square wheel placed on the road with certain points of contact and rotation, and the question of finding the speed of a specific point on the wheel. Through derivatives and observation of a right triangle formed by the points, the speed of the point is found to be secant of the angle between the points.
  • #1
Poetria
267
42

Homework Statement



Consider a bumpy road in which each hump has the shape
##y=\cosh(a)-\cosh(x)## for ##-a\leq x \leq a##
where ##y'(a)=-1## so ##a=arcsinh(1)##

L=2
##(x_p(t), y_p(t)) = (t, \cosh(a)-\cosh(t))##
##(x_q(t), y_q(t)) = (t, \cosh(a))##

We place the square wheel onto the bumpy road so that the point C touches the point A when t=0. The wheel rolls along so that the point D lands on the point B. Hence, the size (the length of one side) of the square wheel is fixed at the value found previously.

images_square-wheel.svg
Point A is at the top of the 0th hump, and
B is where the 0th hump and the 1st hump meet.
Q is the center of the wheel.
C is the middle of the side of the square which touches A at t=0
P is the point of contact of the wheel and the road.
At time t=0, the points A, C, and P coincide.
At all time t, the point P lies directly below the point Q.

images_square-wheel-angle.svg




If the wheel rotates at constant, unit speed, then the angle ##\theta## between P, Q and C satisfies ##\frac {d\theta}{dt} = 1##
For ##0\leq\theta\leq \frac \pi 4## find the speed of Q as a function of t.

2. The attempt at a solution

I have taken the derivatives of x and y: ##(x_q(t), y_q(t)) = (t, \cosh(a))##

##\sqrt{1+\sinh(t)^2}##

but this is wrong. :(
 
Physics news on Phys.org
  • #4
I am not sure what you are asking. The vertical height of point Q above the point of contact must be constant. The speed can be anything.

On edit: Yes, the speed is constant at one unit as stated. Sorry, I missed that.
 
  • Like
Likes Poetria
  • #5
Oh OK. Many thanks. :) The question was about the speed of the point Q.
 
  • #6
Poetria said:
Oh OK. Many thanks. :) The question was about the speed of the point Q.

Well, I have definitely screwed this up. There was a pitfall: dx/dtheta.
 
  • #7
If ##(x_q(t), y_q(t)) = (t, \cosh(a))## then
##\frac{d}{dt}(x_q(t), y_q(t)) = \frac{d}{dt}(t, \cosh(a))=(1, 0)##, no?
 
  • Like
Likes Poetria
  • #8
kuruman said:
If ##(x_q(t), y_q(t)) = (t, \cosh(a))## then
##\frac{d}{dt}(x_q(t), y_q(t)) = \frac{d}{dt}(t, \cosh(a))=(1, 0)##, no?

I thought like you but the correct solution is supposed to be sec(t)! I am trying to understand this. :( :(
 
  • #9
I suspect that the problem is asking to show that if ##\frac{d\theta}{dt}## is constant, the speed of Q is no longer constant. Let me think about this for a while.
 
  • Like
Likes Poetria
  • #10
I think I got it. You have to work with the right triangle ##QCP## and observe that ##QC=1##, ##CP= \sinh(x)## and ##QP= \cosh(x)##. Also observe that ##QC = QP \cos(\theta)##. This is also ##1=\cos(\theta)\cosh(x)##. Take the time derivative (remember the chain rule), do some algebra and you should come up with the answer.
 
  • Like
Likes Poetria
  • #11
Many thanks. :) I have to digest it. :)
 

1. What is a square wheel?

A square wheel is a geometric shape with four equal sides and four right angles. It is often used as a mathematical concept, but can also be found in real-life objects such as toys or vehicles.

2. How does a square wheel move?

A square wheel moves by rolling on its corners, creating a series of smaller, square-shaped movements. This is in contrast to a circular wheel, which rolls smoothly on its circumference.

3. Why is a square wheel not practical for transportation?

A square wheel is not practical for transportation because it creates a very bumpy ride due to the constant change in direction. It also requires a perfectly flat surface to roll on, which is not always possible in real-life situations.

4. Can a square wheel be used to achieve high speeds?

No, a square wheel cannot achieve high speeds because of its constant change in direction. The corners of the square create a lot of friction, making it difficult to maintain a consistent speed.

5. What is the relationship between the speed of the point and the square wheel?

The speed of the point refers to the linear speed of a specific point on the wheel. In the case of a square wheel, the speed of the point is constantly changing due to the shape of the wheel. This makes it difficult to achieve a consistent speed and is one of the reasons why a square wheel is not practical for transportation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
12
Views
934
  • Calculus and Beyond Homework Help
Replies
1
Views
465
  • Calculus and Beyond Homework Help
Replies
5
Views
595
  • Calculus and Beyond Homework Help
Replies
1
Views
997
  • Calculus and Beyond Homework Help
Replies
6
Views
592
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
706
  • Calculus and Beyond Homework Help
Replies
3
Views
765
  • Set Theory, Logic, Probability, Statistics
Replies
0
Views
990
Back
Top