Angular momentum is the product of inertia and its angular velocity. The magnitude of angular momentum of the particle relative to origin is 0.601 \(\bold{kg m^2s^-^1}\).
Angular momentum:
It is defined as the product of inertia and its angular velocity.
The formula for magnitude of angular momentum of a particle,
\(\bold {L = \texttt {mvr sin} \theta}\)
Where,
L - angular momentum
m - mass
v - angular velocityr
r - radius
Radius distance from the origin to particle R =
\(\bold{\sqrt{2^2+ 1^2} = \sqrt{5m} }\)
\(\bold {I = MR^2}\)
\(\bold{I = (0.200)(5)}\\\\\bold{I = 1.00 kgm ^2 }\)
V = 3.0 m/s
V component is at 90° to R
\(\bold { = 3.0(sin 90^o- 63.4^o) }\\\\\bold{= 3.0(sin 26.6^o)}\\\\\bold { = 1.3433 m/s }\)
\(\bold{ \omega =\dfrac{V}{R} \\ = \dfrac{1.3433}{\sqrt{5} } =0.601 rad/s }\)
The magnitude of angular momentum of the particle relative to origin (L)
\(\bold {= I \times \omega }\\\\\bold {= (1.00)(0.601)}\\\\\bold {= 0.601 kgm^2/s}\)
Therefore, the magnitude of angular momentum of the particle relative to origin is 0.601 \(\bold{kg m^2s^-^1}\).
To know more about angular momentum, refer to the link:
https://brainly.com/question/25303285
