How many revolutions per minute would a 15.6 m diameter Ferris wheel need to make for the passengers to feel "weightless" at the topmost point? Do not enter unit.
To determine the number of revolutions per minute required for the passengers of a Ferris wheel to feel "weightless" at the topmost point, we need to consider the concept of centripetal acceleration.
First, let's calculate the velocity of the Ferris wheel at the topmost point. The formula for the linear velocity of an object moving in a circular path is given by:
v = ω * r
where:
v is the linear velocity,
ω (omega) is the angular velocity in radians per second, and
r is the radius of the circular path.
Since we know the diameter of the Ferris wheel, we can calculate its radius by dividing the diameter by 2:
r = 15.6 m / 2 = 7.8 m
Next, we need to determine the required centripetal acceleration at the topmost point. At this point, the passengers should feel weightless, meaning that the net force acting on them is equal to zero. The net force is given by the equation:
F_net = m * a_c
where:
F_net is the net force acting on the passengers,
m is the mass of the passengers, and
a_c is the centripetal acceleration.
Since the passengers should feel weightless, the net force is equal to zero. Therefore, the centripetal acceleration is also equal to zero.
a_c = 0 m/s²
Knowing that centripetal acceleration is calculated as:
a_c = ω² * r
we can rearrange the formula to solve for ω:
ω = √(a_c / r)
Plugging in the given values, we get:
ω = √(0 / 7.8) = 0 rad/s
Therefore, to make the passengers feel weightless at the topmost point, the Ferris wheel would need to rotate at 0 revolutions per minute (rpm).
Note: Please ensure that the units are consistent throughout the calculations. In this case, the radius and diameter should both be in meters.