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A certain solid uniform ball reaches a maximum height h_0 when it rolls up a hill without...

Question:

A certain solid uniform ball reaches a maximum height {eq}h_0 {/eq} when it rolls up a hill without slipping. What maximum height (in terms of {eq}h_0 {/eq}) will it reach if you

(a) double its diameter?

(b) double its mass?

(c) double both diameter and mass?

(d) double its angular speed at the bottom of the hill?

Answer and Explanation:

Let

  • Mass of the ball is m.
  • The maximum height reached by the ball above the bottom of the hill is h.
  • The linear speed of the ball at the bottom of the hill is {eq}v {/eq}.
  • The radius of the ball is r.

Part (a):

As the ball is rolling without slipping so, the angular speed of the ball at the bottom of the hill is;

{eq}\omega = \dfrac{v }{r} {/eq}

The momentum of inertia of the ball is {eq}I = \dfrac{2}{5}mr^2 {/eq}. ( Treating the ball as a solid sphere.)

Potential energy of the ball at the bottom of the hill is {eq}P.E._{bottom} = 0 {/eq}

Kinetic energy of the ball at the top of the hill is {eq}K.E._{top}=0 {/eq}

Potential energy of the ball at the top of the hill is {eq}P.E._{top} = mgh_0 {/eq}

Kinetic energy of the ball at the bottom of the hill is {eq}K.E._{bottom}=\dfrac{1}{2}mv ^2+\dfrac{1}{2}I\omega ^2 {/eq}

From the conservation of mechanical energy, mechanical energy if a ball at the top of the hill is equal to the mechanical energy at the bottom of the hill.

{eq}\begin{align} K.E._{top} +P.E._{top} &= K.E._{bottom} +P.E._{bottom}\\ \implies 0 + mgh_0 &= (\dfrac{1}{2}mv ^2+\dfrac{1}{2}I\omega ^2)+0 \\ mgh_0 &= (\dfrac{1}{2}mv ^2+\dfrac{1}{2}\times (\dfrac{2}{5}mr^2)\times (\dfrac{v }{r})^2)\\ \implies h_0 &= \dfrac{7v ^2}{10g} \\ \end{align} {/eq}

Thus, in this case, the maximum vertical height reached by the ball is;

{eq}h_0 = \dfrac{7v ^2}{10g}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Eqn. 1} {/eq}.

(a) If we double its diameter, the maximum height reached by the ball is;

{eq}h = \dfrac{7v ^2}{10g} = h_0\\ {/eq}.

(b) If we double its mass, the maximum height reached by the ball is;

{eq}h = \dfrac{7v ^2}{10g} = h_0\\ {/eq}.

(c) If we double both diameter and mass, the maximum height reached by the ball is;

{eq}h = \dfrac{7v ^2}{10g} = h_0\\ {/eq}.

(d) If we double its angular speed at the bottom of the hill, the linear velocity at the bottom of the hill will be double and hence the maximum height reached by the ball is;

{eq}h = \dfrac{7(2v) ^2}{10g} =4 h_0\\ {/eq}.


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