You sit at the outer rim of a Ferris Wheel that rotates 2 revolutions per minute (RPM). What would your rotational speed be if y
2 answers:
Answer:
The rotational speed will be the same
Explanation:
When there is a circular motion involved (as in this case with the Ferris Wheel), the rotational speed is defined as the ratio between the angle covered and the time taken to cover that angle (
):
Since all the points along a radius of the the Ferris Wheel rotates coherently, they cover the same angular distance in the same amount of time, so the rotational speed is always the same for every point of the Wheel. This means that if you change the distance by moving halfway from the centre, the rotational speed is still 2 revolutions per minutes, because a point halfway from the centre also covers 2 complete revolutions in a minute.
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Answer:
The resulting, needed force for equilibrium is a reaction from a support, located at 2.57 meters from the heavy end. It is vertical, possitive (upwards) and 700 N.
Explanation:
This is a horizontal bar.
For transitional equilibrium, we just need a force opposed to its weight, thus vertical and possitive (ascendent). Its magnitude is the sum of the two weights, 400+300 = 700 N, since weight, as gravity is vertical and negative.
Now, the tricky part is the point of application, which involves rotational equilibrium. But this is quite simple if we write down an equation for dynamic momentum with respect to the heavy end (not the light end where the additional weight is placed). The condition is that the sum of momenta with respect to this (any) point of the solid bar is zero:
Where momenta from weights are possitive and the opposed force creates an oppossed momentum, then a negative term. Solving our unknown d:
So, the resulting force is a reaction from a support, located at 2.57 meters from the heavy end (the one opposed to the added weight end).
