Question

A solid cylinder (mass 0.332 kg, radius 2.00 cm) rolls without slipping at a speed of 5.00 cm/s. What is its total kinetic energy?

Answer 1

Given data

*The given mass of the solid cylinder is m = 0.332 kg

*The given radius of the solid cylinder is r = 2.00 cm

*The given speed is v = 5.00 cm/s = 0.05 m/s

The formula for the total kinetic energy is given as

`U_T=(1)/(2)I\omega^2+(1)/(2)mv^2`

*Here I is the moment of inertia of the solid cylinder

Substitute the known values in the above expression as

`\begin{gathered} U_T=(1)/(2)((1)/(2)mr^2)*((v)/(r))^2+(1)/(2)mv^2 \\ =(1)/(4)mv^2+(1)/(2)mv^2 \\ =(3)/(4)mv^2 \\ =(3)/(4)(0.332)(0.05)^2 \\ =6.225*10^(-4)\text{ J} \\ =0.6225\text{ mJ} \end{gathered}`