At the instant of the figure, a 7.20 kg particle P has a position vector 7 of magnitude 1.10 m and angle 0, = 40.0° and a velocity vector v of magnitude 1.80 m/s and angle e2 = 31.0°. Force F ,of magnitude 2.70 N and angle 03 = 31.0° acts on P. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle?
At the instant of the figure, a 7.20 kg particle P has a position vector 7 of magnitude 1.10 m and angle 0, = 40.0° and a velocity vector v of magnitude 1.80 m/s and angle e2 = 31.0°. Force F ,of magnitude 2.70 N and angle 03 = 31.0° acts on P. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle?
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter11: Angular Momentum
Section: Chapter Questions
Problem 82AP: A particle has mass 0.5 kg and is traveling along the line x=5.0m at 2.0 m/s In the positive...
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Transcribed Image Text:At the instant of the figure, a 7.20 kg particle P has a position vector of magnitude 1.10 m and angle e, = 40.0° and a velocity
vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque
acting on the particle?
vector v of magnitude 1.80 m/s and angle 02 = 31.0°. Force F ,of magnitude 2.70 N and angle 03 = 31.0° acts on P. All three
0s
02
Units
i
(a) Number
Units
i
(b) Number
Transcribed Image Text:A disk with a rotational inertia of 9.58 kg-m? rotates like a merry-go-round while undergoing a torque given by 7 = (7.18 + 3.15t) N · m.
At time t = 1.00 s, its angular momentum is 7.81 kg-m²/s. What is its angular momentum at t = 3.00 s?
Number
i
16.84
Units
kg-m^2
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