Question

A soccer ball (k = 2/3) with a radius of 0.101 meters and a mass of 0.400 kilograms rolls on a field with an angular speed of 93 radians per second. a. What is the translational kinetic energy of the ball? Include units in your answer.b. What is the rotational kinetic energy of the ball? Include units in your answer.c. The ball then rolls up a hill. What height will the ball reach before coming to rest? Include units in your answer. All answers must be in 3 significant digits.

Answer 1

Given,

The radius of the ball, r=0.101 m

The mass of the ball, m=0.400 kg

The angular speed of the ball, ω=93 rad/s

a.

The linear velocity of the ball is given by,

`v=\omega r_{}`

The translational kinetic energy is given by,

`\begin{gathered} K_t=(1)/(2)mv^2 \\ =(1)/(2)m\omega^2r^2 \end{gathered}`

On substituting the known values,

`\begin{gathered} K_t=(1)/(2)*0.400*93^2*0.101^2 \\ =17.6\text{ J} \end{gathered}`

hus the translational kinetic energy of the ball is 17.6 J

b.

The moment of inertia of the ball is given by,

`I=(2)/(3)mr^2`

The rotational kinetic energy of the ball is given by.

`\begin{gathered} K_r=(1)/(2)I\omega^2 \\ =(1)/(2)*((2)/(3)* mr^2)*\omega^2 \\ =(1)/(3)mr^2\omega^2 \end{gathered}`

On substituting the known values,

`\begin{gathered} K_r=(1)/(3)*0.400*0.101^2*93^2 \\ =11.8\text{ J} \end{gathered}`

hus the trotational kinetic energy of the ball is 11.8 J

.

From the law of conservation of energy, the energy can neither be created nor be destroyed but can be converted from one form to another.

When ball rolls up the hill the ball will lose its translational and rotational kinetic energy. But the lost kinetic energy will be converted into its gravitational potential energy.

Therefore, the gravitational potential energy of the ball when it is up the hill is equal to the sum of its translational and rotational kinetic energy.

Thus,

`\begin{gathered} K_t+K_r=P \\ =\text{mgh} \end{gathered}`

Where g is the acceleration due to gravity and h is the height that the ball will reach before coming to rest.

On rearranging the above equation,

`h=(K_t+K_r)/(mg)`

On substituting the known values,

`\begin{gathered} h=(17.6+11.8)/(0.400*9.8) \\ =7.5\text{ m} \end{gathered}`

herefore the ball will treach a height of 7.5 m before coming to rest.