What is the force experienced by a wheel hitting an edge at 1.5m/s?

  • Thread starter Thread starter Ottoturbo
  • Start date Start date
  • Tags Tags
    Edge Force Wheel
Ottoturbo
Messages
4
Reaction score
0

Homework Statement

Wheel is hitting edge at a x-component speed of 1.5m/s, what is the force it is subjected to when going over the edge. The wheel is loaded with 200N

hjulkant.jpg

Btw, the line with an 60 degree angle to to x-axis/ground is the tangent not a ramp.

Homework Equations



acceleration: speed/time

The Attempt at a Solution


I calculated the acceleration in the y-direction at the distance of 0,052m. Given the x-component speed was constant at 1.5m/s. Time the wheel will use to cover 0,052m: 0,052m divided by 1,5m/s = 0,035s.

The I found the speed in the y-direction: 0,03m/0,035s=0,857m/s
The acceleration this gives: 0,857 m/s divided by 0,035 =24,5 m/s2

Newtons 2. law will give a force of: F=20,4*24,5=500 N

Seems a bit high, and it's only in the y-direction. The force would go though the center of the wheel, so there should be an x-component also.

Is the conservation of angular momentum as the wheel center pivots around the edge the way to go?

What would the moment of inertia be for this wheel? (m*r^2)/2 It's mass is very small, but it's connected to a leg and it's loaded with about 20kgs.
 
Physics news on Phys.org
Welcome to PF!

Hi Ottoturbo! Welcome to PF! :smile:

(have an omega: ω and try using the X2 icon just above the Reply box :wink:)
Ottoturbo said:
Wheel is hitting edge at a x-component speed of 1.5m/s, what is the force it is subjected to when going over the edge. The wheel is loaded with 200N

Is the conservation of angular momentum as the wheel center pivots around the edge the way to go?

no, because the weight has a https://www.physicsforums.com/library.php?do=view_item&itemid=175" isn't conserved … use conservation of energy :smile:

find ω as a function of t …

use that to find the actual acceleration (not the angular acceleration) of the centre of the wheel, explain whether and why the load has the same acceleration, and then use F = (M + m)a to find the force
What would the https://www.physicsforums.com/library.php?do=view_item&itemid=31" be for this wheel? (m*r^2)/2 It's mass is very small, but it's connected to a leg and it's loaded with about 20kgs.

Yes, mr2/2 … the load M isn't rotating (unless it's a washing-machine-type load! :biggrin:), so you don't include M in that moment of inertia formula.

However, don't forget to adjust for the fact that the wheel isn't rotating about its centre of mass.

Also, the load M is rotating (about what?), so you do need to include its angular momentum somehow. :wink:
 
Last edited by a moderator:
So conservation of energy will be: \frac{1}{2}mv_{1}^{2}=\frac{1}{2}mv_{2}^{2}-mgh

\frac{1}{2}\cdot20\cdot1.5^{2}=\frac{1}{2}\cdot20\cdot v_{2}^{2}-20\cdot9.81\cdot0.03v_{2}=1.68 m/s

This gives a higher end speed, top of the egde than to start with, 1.5m/s.

The mass is rotating around the egde. Angular speed would be: \omega=\frac{v}{r}=\frac{1.5}{0.06}=25 \frac{rad}{s}
 
Last edited:
Could I say as the wheel starts to rotate around the edge that the force from the edge must be the same as the centripetal force, F = mv²/r. The wheel has neglectable mass and stops rotating.

F=20*1,5²/0,06=750N
 
Hi Ottoturbo! :smile:
Ottoturbo said:
Could I say as the wheel starts to rotate around the edge that the force from the edge must be the same as the centripetal force, F = mv²/r. The wheel has neglectable mass and stops rotating.

F=20*1,5²/0,06=750N

No … the wheel does not stop rotating, its point of contact remains stationary at the edge while the wheel rotates round it, up to the top level.

And I'm confused :confused: … is the mass or the moment of inertia of the wheel given in the question or not? You did mention it earlier. But if it's not, you needn't bother about it.

The https://www.physicsforums.com/library.php?do=view_item&itemid=27"of the load is v²/r (though v only starts as 1.5, it'll get less), but there's also a tangential acceleration, dv/dt. :wink:

Mass times the total acceleration has to equal the reaction force plus the weight of the load.
 
Last edited by a moderator:
The mass of the wheel is only about 0,1 kg. It's loaded on it's axle with 20kg.
 
ok, then it's probably safe to ignore it.

Use conservation of energy to find v as a function of y, then find v2/r and dv/dt as functions of y. :smile:
 
Back
Top