Question

If a mass of 0.8 kg is constructed at one end of a 0.5m long fiber and rotates in a horizontal circle with an angle of 10 rads, then its angular momentum about the center of rotation, in SI units

Answer 1

Mass, M = 0.8 kg

length of rod, l = 0.5 m

Angle of rotation = 10 rad = 572^o

Radius of circular motion, R = $lcos\theta\space$

As neither the Time period or the linear velocity of the body is mentioned, I am assuming the the time period of rotation to be T_rot

$v=\dfrac{2\pi R}{T_{rot}}\\\Rightarrow v=\dfrac{2\pi\times lcos\theta}{T_{tot}}$

$Angular\space mommentum = mvrsin\theta\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space=\dfrac{m\times2\pi (lcos\theta)^2sin\theta}{T_{tot}}$