# Tire stuck in mud - energy conservation

#### Daniokano

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?

Related Classical Physics News on Phys.org

#### gianeshwar

Without air resistance, show the maximum height snow can reach is h_max=R+v^2/(2g)+gR^2/(2v^2)
Is snow traversing any path?

#### Daniokano

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?

#### willem2

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.

#### Daniokano

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)

#### Daniokano

However this does not account for the gR^2/(2v^2) (given in the solution)

Any ideas?

#### gianeshwar

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.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving