At t = 0 s,a particle of mass 7.08 kg starts from rest at the origin and moves with a velocity
v = 6.00t^2i(hat) − 6.00tj(hat) + 6.00k(hat) m/s.
1. What is the angular momentum of the particle at t = 0 s
I know angular is calculated from moment of inertia * angular velocity but I'm so confused.
at t = 0
v = 6 k
momentum = (7.08*6) k
now to get angular momentum we need a center
I omega = m V cross R = m V x R
it is at the center (R=0) so its angular momentum about the center is zero.
Okay so for t = 0.850 s I would plug in 0.850 for R?
i meant t
To calculate the angular momentum of a particle, we need to use the formula:
Angular Momentum (L) = Moment of Inertia (I) * Angular Velocity (ω)
However, in this case, we are given the particle's velocity, not the angular velocity. So, we need to find the angular velocity first.
The angular velocity (ω) can be obtained from the velocity vector (v) by taking the cross product of the position vector (r) and the velocity vector (v):
Angular Velocity (ω) = r × v
Since the particle starts from the origin (0, 0, 0), the position vector (r) can be represented as:
r = xi(hat) + yj(hat) + zk(hat)
To find the angular velocity, we take the cross product of the position vector (r) and the given velocity vector (v):
ω = r × v
Let's calculate the angular velocity (ω) first:
ω = (xi(hat) + yj(hat) + zk(hat)) × (6.00t^2i(hat) - 6.00tj(hat) + 6.00k(hat))
Expanding the cross product:
ω = (6.00t^2i(hat) - 6.00tj(hat) + 6.00k(hat)) × xi(hat) + (6.00t^2i(hat) - 6.00tj(hat) + 6.00k(hat)) × yj(hat) + (6.00t^2i(hat) - 6.00tj(hat) + 6.00k(hat)) × zk(hat)
Since the cross product of parallel vectors is zero, the above equation can be simplified to:
ω = -6.00t^2j(hat) + 6.00ti(hat)
Now, we can calculate the angular momentum (L) at t = 0s using the given formula:
Angular Momentum (L) = Moment of Inertia (I) * Angular Velocity (ω)
The moment of inertia of a point particle is given by the equation:
Moment of Inertia (I) = mass (m) * distance from the rotation axis squared
In this case, the particle starts from rest at the origin, so the distance from the rotation axis is zero, and the moment of inertia becomes zero.
Therefore, at t = 0s, the angular momentum (L) of the particle is also zero.