Respuesta :
Radius distance from origin to particle = √ (2²+1²) = √5 m = R
I = MR² = (0.200)(5) = 1.00 kg-m²
Θ = arctan 2/1 = 63.4° = R's angle CCW from horizontal
V = 3.0 m/s
V component that is at 90° to R = 3.0(sin 90°- 63.4°) = 3.0(sin 26.6°) = 1.3433 m/s
w = [V component / R] = 1.3433/√5 = 0.601 rad/s
size of angular momentum of particle relative to origin = Iw = (1.00)(0.601) = 0.601 kgm²/s
i hope I'm right
I = MR² = (0.200)(5) = 1.00 kg-m²
Θ = arctan 2/1 = 63.4° = R's angle CCW from horizontal
V = 3.0 m/s
V component that is at 90° to R = 3.0(sin 90°- 63.4°) = 3.0(sin 26.6°) = 1.3433 m/s
w = [V component / R] = 1.3433/√5 = 0.601 rad/s
size of angular momentum of particle relative to origin = Iw = (1.00)(0.601) = 0.601 kgm²/s
i hope I'm right
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 [tex]\bold{kg m^2s^-^1}[/tex].
Angular momentum:
It is defined as the product of inertia and its angular velocity.
The formula for magnitude of angular momentum of a particle,
[tex]\bold {L = \texttt {mvr sin} \theta}[/tex]
Where,
L - angular momentum
m - mass
v - angular velocityr
r - radius
Radius distance from the origin to particle R =
[tex]\bold{\sqrt{2^2+ 1^2} = \sqrt{5m} }[/tex]
[tex]\bold {I = MR^2}[/tex]
[tex]\bold{I = (0.200)(5)}\\\\\bold{I = 1.00 kgm ^2 }[/tex]
V = 3.0 m/s
V component is at 90° to R
[tex]\bold { = 3.0(sin 90^o- 63.4^o) }\\\\\bold{= 3.0(sin 26.6^o)}\\\\\bold { = 1.3433 m/s }[/tex]
[tex]\bold{ \omega =\dfrac{V}{R} \\ = \dfrac{1.3433}{\sqrt{5} } =0.601 rad/s }[/tex]
The magnitude of angular momentum of the particle relative to origin (L)
[tex]\bold {= I \times \omega }\\\\\bold {= (1.00)(0.601)}\\\\\bold {= 0.601 kgm^2/s}[/tex]
Therefore, the magnitude of angular momentum of the particle relative to origin is 0.601 [tex]\bold{kg m^2s^-^1}[/tex].
To know more about angular momentum, refer to the link:
https://brainly.com/question/25303285
