At the instant the displacement of a 2.00 kg object relative to the origin is d
ID: 1558634 • Letter: A
Question
At the instant the displacement of a 2.00 kg object relative to the origin is d = (2.00 m)i + (4.00 m)j - (3.00 m)k, its velocity is v = - (6.10 m/s)i + (2.60 m/s)j + (2.80 m/s)k, and it is subject to a force F = (5.50 N)i - (9.00 N)j + (3.60 N)k. (a) Find the acceleration of the object. (Express your answer in vector form.) d = 2.75i - 4.5j + 1.8k m/s^2 (b) Find the angular momentum of the object about the origin. (Express your answer in vector form.) l = 38i + 25.4j + 59.2k kg middot m^2/s (c) Find the torque about the origin acting on the object. (Express your answer in vector form) tau = -12.6i - 23.7j - 40k N middot m (d) Find the angle between the velocity of the object and the force acting on the object. 125.76 (a) Did you write Newton's second law for the motion? (b) 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? (c) Torque is the cross product of a position vector (extending from a chosen point to the particle) and a force vector. (d) Do you recall how to use a dot product to find the angle between the directions of two vectors?Explanation / Answer
from newton's second law
F = m*a
acceleraion a = F/m
a = 2.75i - 4.50 j + 1.80 k m/s^2
=================
(b)
linear momentum p = mv = 2*( -6.1 i + 2.6j + 2.8 k)
p = -12.2 i + 5.2 j + 5.6 k
angular momentum L = d x P
L = (2i + 4j - 3k ) x (-12.2 i + 5.2 j + 5.6 k)
L = 38.0 i + 25.4 j + 59.2 k kg m^2/s
======================
(c)
torque = d x F
torque = (2i + 4j - 3k ) x (5.50i - 9.00 j + 3.60 k)
torque = -12.6 i - 23.7 j - 40.0 k
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(d)
costheta = (F.v)/(F*v)
F.v = (5.50i - 9.00 j + 3.60 k) . ( -6.1 i + 2.6j + 2.8 k)
F.v = -(5.5*6.1) - (9*2.6) + (3.6*2.8) = -46.87
F = sqrt(5.5^2+9^2+3.6^2) = 11.14
v = sqrt(6.1^2+2.6^2+2.8^2) = 7.20
costheta = -46.87/(11.14*7.20)
theta = 125.76 degrees
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