Question

A cylinder of mass 5 kg and radius 4 m is rolling purely on a horizontal surface. What will be the angular momentum of the cylinder with respect to any point on the ground if its centre of mass moves with the velocity of 20 m/s?

• A. $800\frac{Kg-m^2}{s}$
• B. $600\frac{Kg-m^2}{s}$
• C. $1000\frac{Kg-m^2}{s}$
• D. $400\frac{Kg-m^2}{s}$

Answer 1

The correct option is B $600\frac{Kg-m^2}{s}$

As explained in the previous videoclip,
$L=I\omega +mR{V}_{cm}$
$=\frac{m{R}^2}{2}\omega +mR{V}_{cm}$
also, V = R$\omega$
$\Rightarrow L=\frac{5(4{)}^{2}5}{2}+5.20\times 4$
$=200+400=600\frac{kg-m^2}{s}$