Transcription of PHYSICS 111 HOMEWORK SOLUTION #10
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PHYSICS 111 HOMEWORK SOLUTION #10
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).
Facing the axle, a 22.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round? 11. Without the child the merry-go-round has a moment of inertia I which will change to I0= I+ mr2 when the child hops onto the edge. However, the moment of inertia should be conserved.
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