Question

A solid cylinder (mass 0.274 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 = 0.274 kg

*The given radius of the cylinder is r = 2.00 cm = 0.02 m

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

The formula for the total kinetic energy is given as

`\begin{gathered} U_T=U_k+U_R \\ U_T=(1)/(2)mv^2+(1)/(2)I\omega^2 \\ =(1)/(2)mv^2+(1)/(2)((1)/(2)mr^2)((v)/(r))^2 \end{gathered}`

*Here U_K is the translation kinetic energy

*Here U_R is the rotational kinetic energy

*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)(0.274)(0.05)^2+(1)/(2)((1)/(2)*0.274*(0.02)^2)((0.05)/(0.02))^2 \\ =0.000342+0.000171 \\ =5.13*10^(-4)\text{ J} \\ =5.13*10^(-1)\text{ mJ} \end{gathered}`

Hence, the total kinetic energy is U_T = 5.13 × 10^-1 mJ