# What linear speed must a 0.0518-kg hula hoop have if its total kinetic energy is to be 0.132 J?...

## Question:

What linear speed must a 0.0518-kg hula hoop have if its total kinetic energy is to be 0.132 J? Assume the hoop rolls on the ground without slipping.

## Rotational and translational kinetic energy.

if a ball is kicked, work is done on it, and energy is transformed into kinetic energy, which manifests itself in two ways. Rotational energy related to the rotation of the ball with respect to an axis that passes through its center of mass and translational energy related to the change of position of the ball. To solve the following exercise it is necessary to apply the concepts of kinetic energy, angular and linear speed, in addition to the necessary calculation and algebra tools.

We are given:

• The mass is {eq}m=0.0518\;\text{Kg} {/eq}
• The total kinetic energy is {eq}K_{T}= 0.132\;\text{J} {/eq}
• The radius of the hula hoop is R"'

The moment of inertia of a hula hoop is.

{eq}\displaystyle I=m\;R^{2} {/eq}

Where.

• m is the mass.

The total kinetic energy equals

{eq}\displaystyle K_{T}=K_{rot}+K_{trans} {/eq}

Where.

• Krot is rotational kinetic energy.
• Ktrans is the translational kinetic energy.

Rotational kinetic energy is given by.

{eq}\displaystyle K_{rot}=\frac{1}{2}\;I\;w^{2} {/eq}

Where.

• I is the moment of inertia.
• w is the angular speed.

The translational kinetic energy is given by.

{eq}\displaystyle K_{trans}=\frac{1}{2}\;m\;V^{2} {/eq}

Where.

• m is the mass.
• V is linear speed.

Replacing in the expression of energy.

{eq}\displaystyle K_{T}=K_{rot}+K_{trans} \\ \displaystyle K_{T}=\frac{1}{2}\;I\;w^{2}+\frac{1}{2}\;m\;V^{2} {/eq}

Now, the relationship between linear speed and angular speed is.

{eq}\displaystyle w=\frac{V}{R} {/eq}

Where.

• V is linear speed.

Replacing the expression of inertia and angular speed.

{eq}\displaystyle K_{T}=\frac{1}{2}\;I\;w^{2}+\frac{1}{2}\;m\;V^{2} \\ \displaystyle K_{T}=\frac{1}{2}\;m\;R^{2}\;(\frac{V}{R})^{2}+\frac{1}{2}\;m\;V^{2}\\ {/eq}

Simplified and obtaining the expression for linear speed.

{eq}\displaystyle K_{T}=\frac{1}{2}\;m\;R^{2}\;(\frac{V}{R})^{2}+\frac{1}{2}\;m\;V^{2}\\ \displaystyle K_{T}=\frac{1}{2}\;m\;R^{2}\;\frac{V^{2}}{R^{2}}+\frac{1}{2}\;m\;V^{2}\\ \displaystyle K_{T}=\frac{1}{2}\;m\;V^{2}+\frac{1}{2}\;m\;V^{2}\\ \displaystyle K_{T}=\frac{2}{2}\;m\;V^{2}\\ \displaystyle K_{T}=m\;V^{2}\\ \displaystyle m\;V^{2}=K_{T}\\ \displaystyle V^{2}=\frac{K_{T}}{m}\\ {/eq}

Finally

{eq}\displaystyle V=\sqrt{\frac{K_{T}}{m}}\\ {/eq}

Numerically evaluating.

{eq}\displaystyle V=\sqrt{\frac{K_{T}}{m}}\\ \displaystyle V=\sqrt{\frac{0.132\;\text{J}}{0.0518\;\text{Kg}}}\\ \displaystyle V=1.60\;\text{m}/\text{sec} {/eq}

Finally:

The linear speed is {eq}\displaystyle \color{red}{\boxed{V=1.60\;\text{m}/\text{sec}}} {/eq}