Transcription of PHYSICS 111 HOMEWORK SOLUTION #10 - New Jersey …
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PHYSICS 111 HOMEWORK SOLUTION #10 - New Jersey …
PHYSICS 111 HOMEWORK . SOLUTION #10. April 10, 2013. Given M~ = 4~i + ~j 3~k and N. ~ = ~i 2~j 5~k, calculate the vector ~ ~. product M N . By simply following the rules of the cross product: ~i ~i = ~j ~j = ~k ~k = ~0. ~i ~j = ~k = ~j ~i ~j ~k = ~i = ~k ~j ~k ~i = ~j = ~i ~k ~ N. M ~ = (4~i + ~j 3~k) (~i 2~j 5~k). = 8~k + 20~j ~k 5~i 3~j 6~i = 11~i + 17~j 9~k Calculate the net torque (magnitude and direction) on the beam in the figure below about the following axes. 2. We will choose clockwise as our positive direction and apply the formula for a torque: X. ~ net = F~i ~ri X. net = Fi ri sin i a) About the O-axis: net = 25 2 sin 60 + 10 4 sin 20 + 0. = This net torque is counterclockwise b) About the C-axis: net = 0 + 10 2 sin 20 30 2 sin 45. = This net torque is again counterclockwise A light, rigid rod of length l = m joins two particles, with masses m1 = kg and m2 = kg, at its ends.
Earth as a solid sphere will have a moment of inertia of 2 5 MR, it’s mean radius R is about 6378 Km, with a mass M of 5.9736 1024Kg, While spinning it makes a whole revolution in 24 hours. Angular momentum due to spinning is thus, L = I! = 2 5 MR2! = 2 5 5:9736 1024 63780002 2 24 3600 = 7:07 1033 kg:m2=s Spinning is about the north celestial ...
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