Asked by Brittany
A turntable of radius 2m rotates freely about a fixed vertical axis. A child 19 kg is standing on the turntable right at the outer edge and the system is initially rotating at an angular speed of 0.5 rev/s. The child then moves to position 1m from center of the turntable. what does the urntable angular speed become? The moment of inertia is 400kgm^2.
Answers
Answered by
bobpursley
conservation of momentum
29*2^2*.5+400*.5=29*1^2*w + 400*w
solve for w, in revs/sec
29*2^2*.5+400*.5=29*1^2*w + 400*w
solve for w, in revs/sec
Answered by
Brittany
I don't really understand we use I1w1=I2w2?
but according to your equation its
mr^2+Iw = mr^2+Iw ... which is equivalent to I+L = I+L <-- i don't get this part?
and the answer is 0.568 rev/s which according to your way is right but its 19 instead of 29.
but according to your equation its
mr^2+Iw = mr^2+Iw ... which is equivalent to I+L = I+L <-- i don't get this part?
and the answer is 0.568 rev/s which according to your way is right but its 19 instead of 29.
Answered by
Brittany
i mean..
mr^2w+Iw = mr^2w+Iw ... which is equivalent to L+L = L+L
mr^2w+Iw = mr^2w+Iw ... which is equivalent to L+L = L+L
Answered by
bobpursley
The moment of inertialfor the kid is mass*distance^2.
sorry about the 29.
sorry about the 29.
Answered by
Brittany
ohh okies thanks for clearing that up (:
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