At a particular instant, a 1.0 kg particle's position is r = (4.015.0ĵ+ 2.0k) m, its velocity is ✔ = (-1.01 + 5.0ĵ+ 1.0k) m/s, and the force on it is F = (13.01 + 17.0ĵ) N. (Express your answers in vector form.) (a) What is the angular momentum (in kg. m²/s) of the particle about the origin? 1 = kg - m²/s (b) What is the torque (in N m) on the particle about the origin? T= N.m (c) What is the time rate of change of the particle's angular momentum about the origin at this instant (in kg - m²/s²)? dī dt kg - m²/s² =
At a particular instant, a 1.0 kg particle's position is r = (4.015.0ĵ+ 2.0k) m, its velocity is ✔ = (-1.01 + 5.0ĵ+ 1.0k) m/s, and the force on it is F = (13.01 + 17.0ĵ) N. (Express your answers in vector form.) (a) What is the angular momentum (in kg. m²/s) of the particle about the origin? 1 = kg - m²/s (b) What is the torque (in N m) on the particle about the origin? T= N.m (c) What is the time rate of change of the particle's angular momentum about the origin at this instant (in kg - m²/s²)? dī dt kg - m²/s² =
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Transcribed Image Text:At a particular instant, a 1.0 kg particle's position is r = (4.0î – 5.0ĵ + 2.0k) m, its velocity is ✔ = (−1.0Î + 5.0ĵ + 1.0k) m/s, and the force on it is F = (13.0î + 17.0ĵ) N. (Express your answers in
vector form.)
-
(a) What is the angular momentum (in kg · m²/s) of the particle about the origin?
ī
kg . m²/s
=
(b) What is the torque (in N·m) on the particle about the origin?
+
T =
(c) What is the time rate of change of the particle's angular momentum about the origin at this instant (in kg • m²/s²)?
dī
dt
N•m
=
2
kg. m²/s²
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