Question

A solid sphere rolling without slipping on a horizontal surface. If the translational speed of the sphere is 2.00 m/s, what is its total kinetic energy?

Answer 1

The total kinetic energy is 2.8m J. (NOTE: m is mass of the sphere)

Step-by-step explanation:

The total kinetic energy of a sphere is given by the sum of the rotational kinetic energy and the translational kinetic energy. That is,

`K_(Total) = K_(R) + K_(T)`

The rotational kinetic energy
`K_(R)` is given by

`K_(R) = (1)/(2)I\omega^(2)`

Where
`I` is the moment of inertia

and
`\omega` is the angular velocity

The translational kinetic energy
`K_(T)` is given by

`K_(T) = (1)/(2)mv^(2)`

Where
`m` is the mass

and
`v` is the translational speed (velocity)

∴
`K_(Total) = (1)/(2)I\omega^(2) + (1)/(2)mv^(2)`

But, the moment of inertia
`I` of a sphere is given by

`I = (2)/(5)mr^(2)`

Where
`m` is mass

and
`r` is radius

∴
`K_(Total) = (1)/(2)* (2)/(5)mr^(2) \omega^(2) + (1)/(2)mv^(2)`

`K_(Total) = (1)/(5)mr^(2) \omega^(2) + (1)/(2)mv^(2)`

Also,
`\omega = (v)/(r)`

∴
`\omega^(2) = (v^(2) )/(r^(2) )`

Then,

`K_(Total) = (1)/(5)mr^(2) * (v^(2) )/(r^(2) ) + (1)/(2)mv^(2)`

`K_(Total) = (1)/(5)mv^(2) + (1)/(2)mv^(2)`

∴
`K_(Total) = (7)/(10)mv^(2)`

From the question,
`v = 2.00 m/s`

Then,

`K_(Total) = (7)/(10)m(2.00)^(2)`

`K_(Total) = (7)/(10)m* 4.00`

`K_(Total) = 2.8m J`

Hence, the total kinetic energy is 2.8m J. (NOTE: m is mass of the sphere)