Question

How many revolutions per minute would a 26 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the topmost point?

Answer 1

Answer:

Explanation:

Let m be the mass of passenger.

diameter of wheel, d = 26 m

radius of wheel, r = half of diameter = 13 m

Let ω be the angular velocity of the Ferris wheel.

A the passengers becomes weightless, so the centripetal force acting on the passengers is balanced by the weight of passengers.

mg = m r ω²

9.8 = 12 x ω²

ω = 0.9 rad/s

Let f be the frequency

ω = 2 π f

0.9 = 2 x 3.14 x f

f = 0.143 revolutions per second

Number of revolutions per minute = 0.143 x 60

                                                         = 8.6 revolutions per minute

Answer 2

Answer:$N=8.28\ rpm$

Explanation:

Given

Diameter of wheel $d=26\ m$

Person is feeling Weightlessness i.e. Net force on the person is equivalent to its weight

At top point weight is equal to Centripetal force

$mg=\frac{mv^2}{r}$

where v=velocity of wheel

thus

$g=\frac{v^2}{R}$

$v=\sqrt{gR}$

$v=\sqrt{9.8\times 13}$

$v=11.28\ m/s$

$v=\frac{\pi d\cdot N}{60}$

$11.28=\frac{\pi \cdot 26\cdot N}{60}$

$N=8.28\ rpm$