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

Step-by-step 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`

Step-by-step 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=(mv^2)/(r)`

where v=velocity of wheel

thus

`g=(v^2)/(R)`

`v=√(gR)`

`v=√(9.8* 13)`

`v=11.28\ m/s`

`v=(\pi d\cdot N)/(60)`

`11.28=(\pi \cdot 26\cdot N)/(60)`

`N=8.28\ rpm`