A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the acceleration of a passenger at his or her lowest point during the ride?
2007-12-06 16:18:38 · 5 answers · asked by roncho 4 in Science & Mathematics ➔ Physics
The tangential acceleration is found from the equation
a = r ω²
r = radius in meters
ω = angular velocity in radians / second
Your angular velocity is five turns per minute = ( 2π )( 5 ) / ( 60 ) radians per second.
At the bottom of the wheel, the net acceleration is the above ( up ) - g ( down )
a = r ω² - g
where g = the local acceleration of gravity, 9.8 m/s^2
2007-12-06 16:27:06 · answer #1 · answered by jgoulden 7 · 1⤊ 0⤋
Frequency= number of turns per second = 5/(60X1)=1/12 Hz.
acceleration= W^2XRadius.
acceleration=(2(pi)F)X(15)
acceleration=(2(pi)(1/12)X(15)
acceleration=7.853981634 ms^-2
2007-12-07 00:28:43 · answer #2 · answered by Murtaza 6 · 0⤊ 0⤋
acceleration = 0; since the rotational velocity is a constant (5rpm), the acceleration = 0 unless the ferris wheel is starting or stopping
2007-12-07 00:26:01 · answer #3 · answered by Hectorux 2 · 0⤊ 0⤋
I was in higher level physics in junior and senior year. This advanced program called IB.
But I didnt' like physics much. Or school.
So, I spent the entire allotted time writing a poem on my HL physics test.
They gave me a 3 / 7 lol.
Sorry for my lame answer. But it was related. :)
2007-12-07 00:21:03 · answer #4 · answered by whimsy 3 · 0⤊ 0⤋
do u even know the answer?
2007-12-07 00:21:25 · answer #5 · answered by Anonymous · 0⤊ 0⤋