A Ferris wheel with radius 9.8 m rotates at a constant rate, completing one revolution every 36.6 s. Suppose the Ferris wheel begins to decelerate at the rate of 0.227 rad/s2 when a passenger is at the top of the wheel. Find the magnitude and direction of the passenger's acceleration at that time.
_____magnitude
direction______° below direction of travel
first find angular speed
w=(2pi)/36.6 to get rad/s
then, you must find a(cp) and a(tangential)
acp=r*w2
at=r*alpha =>9.8*-.227
use a=sqrt(acp^2+at^2)
the answer is the magnitude
to find the angle use tan^-1(acp/at)
To find the magnitude and direction of the passenger's acceleration, we can use the concepts of circular motion and centripetal acceleration.
First, let's find the angular speed of the Ferris wheel. We know that it completes one revolution (2π radians) every 36.6 seconds. Therefore, the angular speed can be calculated as:
Angular speed (ω) = (2π radians) / (time taken for one revolution)
= (2π radians) / (36.6 s)
Next, let's find the magnitude of the passenger's acceleration. The acceleration of an object moving in a circle at a constant speed is given by:
Acceleration (a) = (angular speed)² * radius
So, we can substitute the values into the equation:
Acceleration = (angular speed)² * radius
= (ω)² * r
where r is the radius of the Ferris wheel, given as 9.8 m.
Now, let's calculate the magnitude:
Magnitude of passenger's acceleration = (ω)² * r
To find the direction of the passenger's acceleration, we know that it is directed towards the center of the circular motion (towards the bottom of the wheel). At the top of the wheel, the acceleration will be directed downwards.
So, the direction of the passenger's acceleration is towards the center of the wheel or downwards.
In summary:
Magnitude of passenger's acceleration = (ω)² * r
Direction of passenger's acceleration = downwards or towards the center of the wheel