To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ
sec θ = 1 / cos <span>θ
csc </span>θ = 1 / sin θ<span>
Using these identities:
</span>cot θ ∙ sec θ = (cos θ / sin θ ) (<span> 1 / cos </span><span>θ)
</span>
We can cancel out cos <span>θ, leaving us with
</span>cot θ ∙ sec θ = 1 / sin θ
cot θ ∙ sec θ = = csc <span>θ</span>
Answer:
Approximately 
, assuming that the gravitational field strength is 
.
Explanation:
Let 
denote the required angular velocity of this Ferris wheel. Let 
denote the mass of a particular passenger on this Ferris wheel.
At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:
- Weight of the passenger (downwards),
, and possibly - Normal force
that the Ferris wheel exerts on this passenger (upwards.)
This passenger would feel "weightless" if the normal force on them is 
- that is, 
.
The net force on this passenger is 
. Hence, when , the net force on this passenger would be equal to .
Passengers on this Ferris wheel are in a centripetal motion of angular velocity around a circle of radius 
. Thus, the centripetal acceleration of these passengers would be 
. The net force on a passenger of mass would be 
.
Notice that 
. Solve this equation for , the angular speed of this Ferris wheel. Since and 
:
.
.
The question is asking for the angular velocity of this Ferris wheel in the unit 
, where 
. Apply unit conversion:
.
No
Imagine your hold a cube (imagine the fists are just hands pushing and the face is the box) it will not move as the are evening each other out
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