At the instant of the figure, a 1.6 kg particle P has a position vector 7 of magnitude 1.7 m and angle 01 = 45° and a velocity vector v of magnitude 3.8 m/s and angle 02 = 30°. Force of magnitude 1.6 N and angle 03 = 30° acts on P. All three vectors lie in the xy plane. (Express your answers in vector form.) (a) What is the angular momentum of the particle about the origin? I = 5.168 kg · m2/s (b) What is the torque acting on the particle about the origin? * = 1.4z N. m Angular momentum is the cross product of a position vector (extending from a chosen point, here the origin, to the particle) and a momentum vector. Do you recall that momentum is p = mv? Do you cross product in magnitude-angle notation, using a right-hand rule to find the direction? You need the angle between the directions of the two vectors. Is that the angle given? recall how to take Torque is the cross product of a position vector (extending from a chosen point, here the origin, to the particle) and a force vector. The right-hand rule of cross products gives the direction.
At the instant of the figure, a 1.6 kg particle P has a position vector 7 of magnitude 1.7 m and angle 01 = 45° and a velocity vector v of magnitude 3.8 m/s and angle 02 = 30°. Force of magnitude 1.6 N and angle 03 = 30° acts on P. All three vectors lie in the xy plane. (Express your answers in vector form.) (a) What is the angular momentum of the particle about the origin? I = 5.168 kg · m2/s (b) What is the torque acting on the particle about the origin? * = 1.4z N. m Angular momentum is the cross product of a position vector (extending from a chosen point, here the origin, to the particle) and a momentum vector. Do you recall that momentum is p = mv? Do you cross product in magnitude-angle notation, using a right-hand rule to find the direction? You need the angle between the directions of the two vectors. Is that the angle given? recall how to take Torque is the cross product of a position vector (extending from a chosen point, here the origin, to the particle) and a force vector. The right-hand rule of cross products gives the direction.
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Transcribed Image Text:At the instant of the figure, a 1.6 kg particle P has a position vector r of magnitude 1.7 m and angle 01
= 45° and a velocity vector v of magnitude 3.8 m/s and angle 02
= 30°. Force F of magnitude
1.6 N and angle 03 = 30° acts on P. All three vectors lie in the xy plane. (Express your answers in vector form.)
(a) What is the angular momentum of the particle about the origin?
Z = 5.168
kg • m2/s
(b) What is the torque acting on the particle about the origin?
T = 1.4z
N: m
Angular momentum is the cross product of a position vector (extending from a chosen point, here the origin, to the particle) and a momentum vector. Do you recall that momentum is p = mv? Do you
recall how to take a cross product in magnitude-angle notation, using a right-hand rule to find the direction? You need the angle between the directions of the two vectors. Is that the angle given?
Torque is the cross product of a position vector (extending from a chosen point, here the origin, to the particle) and a force vector. The right-hand rule of cross products gives the direction.
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