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. The combination rotates in the xy plane about a pivot through the center of the rod (see figure below).

If the two rockets supply a thrust of F=110 N each, we can write Newton 2nd law as follows: 2Fr = I = mr2 = 2F mr The average time interval is then calculated from and != p g r: t =! = mr 2F r g r = m p rg 2F = 8680 s 0.8 A playground merry-go-round of radius R = 1.60 m has a moment of inertia I = 255 kg :m2 and is rotating at 9.0 rev/min about a

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