At the instant the displacement of a 2.00 kg object relative to the origin is Mo
ID: 1285848 • Letter: A
Question
At the instant the displacement of a 2.00 kg object relative to the origin is ModifyingAbove k With ? components respectively), and (j) the angle between the velocity of the object and the force acting on the object. ModifyingAbove j With ? and ModifyingAbove i With ? , ModifyingAbove k With ? components respectively), the torque about the origin acting on the object ((g), (h) and (i) for ModifyingAbove j With ? and ModifyingAbove i With ? , ModifyingAbove k With ? components respectively), the angular momentum of the object about the origin ((d), (e) and (f) for ModifyingAbove j With ? and ModifyingAbove i With ? , ModifyingAbove k With ?. Find the acceleration of the object ((a), (b) and(c) for ModifyingAbove j With ? + (6.39 N) ModifyingAbove i With ? - (1.06 N) ModifyingAbove Upper F With right-arrow = (6.64 N) ModifyingAbove k With ? and it is subject to a force ModifyingAbove j With ? + (7.18 m/s) ModifyingAbove i With ? + (5.95 m/s) ModifyingAbove v With right-arrow = - (3.87m/s) ModifyingAbove k With ?, its velocity is ModifyingAbove j With ? - (3.00 m) ModifyingAbove i With ? + (4.00 m) ModifyingAbove d With right-arrow = (2.00 m)Explanation / Answer
Find the acceleration of the object.
acceleration, 'a' = F/m = (3.32m/s^2)i - (0.53m/s^2)j + (3.19 m/s^2)k
angular momentum =
r x p
r = 2.00,i + 4.00j, - 3.00k m
p = m*v = - 7.74i+ 11.9j + 14.36kg*m/s
now from cross product we get 21.74i - 51.94j + 54.76k
Torque T = r x F
r = 2.0i + 4.00j, - 3.00k m
F = 6.64i-1.06j+6.39k
take the cross product of these vectors.
28.74i+7.14j-28.68k
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