Hanging beads in vehicle turning on horizontal road (constant v and r)

In summary, the car has two strings, each supporting a bead. The tension in the strings resolves into m*v^2/r horizontally and mg vertically. The car moves along a curved path, and the tension in the strings affects the angular velocities of the beads.
  • #1
AJG3
4
1
TL;DR Summary
One hanging bead (massless string) in a turning car would deflect to angle atan(v^2/r*g) from vertical, but with a straight string, multiple beads are each at a different radius. So what is the curve of the string?
Neglect any forces within string, except tension.
 
Physics news on Phys.org
  • #2
Where are you stuck?
You must show your attempts at an answer before we can help you.
 
  • #3
Welcome @AJG3 !
Also, could you show us a diagram showing the meaning of "but with a straight string, multiple beads are each at a different radius."
 
  • Like
Likes nasu
  • #4
Lnewqban said:
Welcome @AJG3 !
Also, could you show us a diagram showing the meaning of "but with a straight string, multiple beads are each at a different radius."
Pretty sure he means:
  • a bead at 2 inches along the string has a rotational radius of 2 inches
  • a bead at 4 inches along the string has a rotational radius of 4 inches
  • etc.
 
  • Like
Likes Lnewqban
  • #5
For specificity, I assumed n point mass beads, each with 1 unit of string length between. The last (outermost) will be bead 1, next inwards bead 2, and so forth I immediately realized that not only is each bead at a different radius, but also a different velocity, ω*ri, with ω the angular velocity of the car and all the beads. Let v and r apply to the outermost bead then ω = v/r.

The string supporting the outermost bead is at angle atan(v^2/rg) from vertical to balance forces including centripetal force. The tension in the string resolves into m*v^2/r horizontally and mg vertically.

Moving on to bead 2:
r2 = r - sin(atan(v^2/rg) = r - (v^2/rg)/√(1+(v^2/rg)^2)
the vertical component of T supporting bead 2 must = 2*mg
the horizontal component of T supporting bead 2 must be mv2^2/r2+mv^2/r =
mω^2r2+mω^2r = mω^2(2r - (v^2/rg)/√(1+(v^2/rg)^2)))

so each successive centripetal force on the bead gets a bit stronger and the angles a bit farther from vertical, but beyond writing the second angle as an atan, I'm not seeing how to get the overall curve. It looks to me that the algebra explodes without major simplification.

Perhaps the answer is a continuous solution with no beads and distributed mass in the string. Since I always fail at deriving the catenary, I'm sure that problem would stump me too.
 
  • #6
DaveC426913 said:
Pretty sure he means:
  • a bead at 2 inches along the string has a rotational radius of 2 inches
  • a bead at 4 inches along the string has a rotational radius of 4 inches
  • etc.
Sorry. I can't draw a figure, but what I meant is that unless the beads are hanging vertically (they won't in a turning car), then each is at a different radius from the center of the car's circular path. Obviously, a sensible person makes the simplification that all the beads are within inches of each other and the car's turning radius is hundreds of feet, so the beads all have approximately the same turning radius and hang in a straight inclined line. But that is not the problem I'm trying to solve.
 
  • #7
AJG3 said:
Sorry. I can't draw a figure, but what I meant is that unless the beads are hanging vertically (they won't in a turning car), then each is at a different radius from the center of the car's circular path.
Something like this?

Turning car beads.jpg
 
  • #8
Thank you. That is it - looking forward for a left turn. And I realize I was wrong and the upper string would indeed be deflected less from vertical than the lower string: twice the gravitational force, but less than twice the centripetal force, since although it’s r is smaller, it’s v^2 is smaller still. mv^2/r = m* ω^2*r^2/r. So directly, not inversely proportional to r.

Reference: https://www.physicsforums.com/threa...l-road-constant-v-and-r.1047384/#post-6822920 ^2*r^2/r, so directly, not inversely, proportional to r.
 
  • Like
Likes Lnewqban
  • #9
AJG3 said:
Perhaps the answer is a continuous solution with no beads and distributed mass in the string. Since I always fail at deriving the catenary, I'm sure that problem would stump me too.
Yes, same for me.

My inclination would be to give up and go with a numerical approach for the finite case. Make an initial guess for a set of beads (say the straight line approximation) and compute the net force on each bead. Then relax the solution so that each bead moves in the direction of the net force, subject to the constraints imposed by the string. Iterate while a stable solution is approached.

This would be the "hill climbing" approach to optimization.

Graph the result and see if one can intuit an equation that fits the curve.
 

Related to Hanging beads in vehicle turning on horizontal road (constant v and r)

1. How do hanging beads in a vehicle turning on a horizontal road demonstrate centripetal force?

When a vehicle turns on a horizontal road, the hanging beads will appear to move towards the outside of the turn, away from the center of the circle. This is due to the centripetal force, which is the force that keeps an object moving in a circular path. In this case, the centripetal force is provided by the tension in the string holding the beads, which pulls them towards the center of the circle.

2. What is the relationship between the speed of the vehicle and the distance between the beads when turning?

The distance between the beads will decrease as the speed of the vehicle increases. This is because the centripetal force required to keep the beads moving in a circular path is directly proportional to the square of the speed. As the speed increases, the centripetal force also increases, causing the beads to move closer together.

3. How does the radius of the turn affect the movement of the hanging beads?

The radius of the turn directly affects the centripetal force required to keep the beads moving in a circular path. As the radius decreases, the centripetal force also decreases, causing the beads to move further away from the center of the circle. This can be observed by the beads appearing to spread out when the vehicle turns on a larger radius compared to a smaller radius.

4. Can the hanging beads be used to determine the speed of the vehicle?

Yes, the hanging beads can be used to estimate the speed of the vehicle. By measuring the distance between the beads and knowing the radius of the turn, the speed of the vehicle can be calculated using the formula v = √(r * g * tanθ), where v is the speed, r is the radius, g is the acceleration due to gravity, and θ is the angle of the string with the vertical.

5. How does the angle of the string holding the beads affect their movement when turning?

The angle of the string with the vertical affects the magnitude of the centripetal force acting on the beads. The greater the angle, the larger the centripetal force, and therefore the closer the beads will be to the center of the circle. This can be observed by changing the angle of the string and noting how the beads move closer or further away from the center of the circle when the vehicle turns.

Similar threads

Replies
3
Views
792
  • Mechanical Engineering
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
339
Replies
13
Views
1K
Replies
37
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
889
Replies
4
Views
2K
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
861
Back
Top