Question

The drawing shows a tire of radius R on a moving car

Answer 1

Answer:

 H / R = 2/3

Explanation:

Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.

Initial. Lowest point

          Em₀ = K = 1/2 m v²

Final. In the sought height

         $Em_{f}$ = U = mg h

Energy is conserved

        Em₀ =  $Em_{f}$

        ½ m v² = m g h

        v² = 2 gh

Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass

Initial. Lower

        Em₀ = K = ½ I w²

Final. Heights sought

        Emf = U = m g R

        Em₀ =  $Em_{f}$

       ½ I w² = m g R

The moment of inertial of a cylinder is

       I = $I_{cm}$ + ½ m R²

      I= ½ $I_{cm}$  + ½ m R²

Linear and rotational speed are related

       v = w / R

       w = v / R

We replace

      ½ $I_{cm}$  w² + ½ m R² w²  = m g R

moment of inertia of the center of mass      

      $I_{cm}$  = ½ m R²

     ½ ½ m R² (v²/R²) + ½ m v² = m gR

    m v² ( ¼ + ½ ) = m g R

   

       v² = 4/3 g R

As they indicate that the linear velocity of the two points is equal, we equate the two equations

        2 g H = 4/3 g R

         H / R = 2/3