Tire stuck in mud - energy conservation

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Discussion Overview

The discussion revolves around a physics problem involving a car tire stuck in mud and the maximum height that mud can reach when splashed from the tire. Participants explore the application of conservation of energy principles and the role of various parameters such as tire radius and velocity in determining the height of the mud.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes using conservation of energy to relate the initial kinetic energy of the tire to the gravitational potential energy of the mud.
  • Another participant questions how the radius of the tire fits into the energy conservation equation.
  • A later reply clarifies the problem statement, emphasizing the need to consider the height at which the mud leaves the tire.
  • One participant suggests an alternative approach without using conservation of energy, deriving equations of motion for the mud after it leaves the tire.
  • Another participant points out that this alternative approach does not account for a specific term (gR^2/(2v^2)) mentioned in the original problem statement.
  • One participant discusses the concept of centripetal force and how it relates to the trajectory of the mud, suggesting that the mud follows a parabolic path once it leaves the tire.
  • Another participant questions the assumption that the height at which the mud leaves the tire is equal to the radius R.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate approach to solving the problem, with some advocating for conservation of energy while others propose kinematic equations. There is no consensus on the correct method or the assumptions regarding the height at which the mud leaves the tire.

Contextual Notes

Assumptions about the initial conditions, such as the height of the mud when it leaves the tire and the effects of forces acting on the mud, remain unresolved. The discussion includes various interpretations of the problem statement and the equations involved.

Daniokano
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Rg. Without air resistance, show the maximum height snow can reach is h_max=R+v^2/(2g)+gR^2/(2v^2)
Solve using conservation of energy

How do I start this problem? I assume all the initial kinetic energy ((mv^2)/2) from the spinning of the tire is translated to the gravitational potential energy (mgh) but how does the radius fit into the equation? Help please?" slated to the gravitational potential energy (mgh) but how does the radius fit into the equation? Help please?
 
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Daniokano said:
Without air resistance, show the maximum height snow can reach is h_max=R+v^2/(2g)+gR^2/(2v^2)
Please express the problem clearly.
Is snow traversing any path?
 
My apologies, I didn't copy the question out correctly :

The problem statement, all known variables and given data
A car is stuck in the mud and mud is splashed around the rim of the tires. Assume that the radius of a tire is R and is spinning at a speed v>Rg. Without air resistance, show the maximum height the mud can reach is h_max=R+v^2/(2g)+gR^2/(2v^2)
Solve using conservation of energy

2.Relevant Equations
Ei=Ef
Ei=(1/2)mv2
Ef=mgh

How do I start this problem? I assume all the initial kinetic energy ((mv^2)/2) from the spinning of the tireis translated to the gravitational potential energy (mgh) using E_i=E_f, but how does the radius fit into the equation?
 
Daniokano said:
How do I start this problem? I assume all the initial kinetic energy ((mv^2)/2) from the spinning of the tireis translated to the gravitational potential energy (mgh) using E_i=E_f, but how does the radius fit into the equation?
You should also consider the height where the mud leaves the tire.
 
willem2 said:
You should also consider the height where the mud leaves the tire.

OK, so without use of conservation of energy, I assumed the height the mud left the tire is R. Now the velocity of the the mud is given v_t and the only acceleration effecting the mud is -g.

y(t) = R + v_t*t - (g*t^2)/2

v(t) = v_t - g*t

a(t) = - g

Max height is reached when v(t) = 0, solve for t.

t = v_t/g

Sub for t to find subsequent height

h_max = R + (v_t)^2/g - (v_t)^2/(2g)

h_max = R + (v_t)^2/(2g)
 
However this does not account for the gR^2/(2v^2) (given in the solution)

Any ideas?
 
When centrepetal force(adhesion) no longer can hold the mud or snow it goes in a parabolic path with the maximum (as required in problem by a part of mud as there are many portions of mud being slung at lower trajectories also) trajectory.It means we need initial velocity of that maximum trajectory and the portion is now free of tyre.Rest is now the usual path of a projectile problem.
 
Daniokano said:
I assumed the height the mud left the tire is R
Why assumed R?
 

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