A Ferris wheel (a vertical circle for the purposes of this question) with a radius 42.2 m is initially rotating at a constant rate, completing one revolution every 33.2 seconds. Suppose the Ferris wheel begins to decelerate at the rate of 0.0542 rad/s2.
[Note: If it helps you, for sake of definiteness, you may take the wheel to be rotating clockwise… though this is immaterial.]
a) Find the magnitude of the passenger's acceleration at that instant in m/s2.
b) What is the angle between the passenger's velocity vector and acceleration vector at that instant (in degrees)?
ω₀=2πn₀=2π/33.2=0.19 rad/s
centripetal acceleration is a(cen) = ω₀²R=0.19²•42.2=1.52 m/s²
tangential acceleration is a(tan) =ε•R =0.0542•42.2=0.12 m/s²
acceleration is a=sqrt{ a(cen)²+a(tan)²}= sqrt{1.52²+0.12²}=1.525
α=arctan(a(tan)/a(cen)} =4.51°
the angle between the passenger's velocity vector and acceleration vector =90+4.51 =94.51°
Thank you very much!
To solve this problem, we need to use the kinematic equations for rotational motion.
a) To find the magnitude of the passenger's acceleration, we can use the equation:
a = r * α
Where a is the acceleration, r is the radius of the Ferris wheel, and α is the angular acceleration.
Given that the radius of the Ferris wheel is 42.2 m and the angular acceleration is -0.0542 rad/s^2 (deceleration is negative), we can substitute these values into the equation to calculate the magnitude of the acceleration:
a = (42.2) * (-0.0542)
a ≈ -2.2864 m/s^2
Therefore, the magnitude of the passenger's acceleration at that instant is approximately 2.2864 m/s^2.
b) The angle between the passenger's velocity vector and acceleration vector can be found using the formula:
θ = arccos(a / v)
Where θ is the angle between the vectors, a is the acceleration, and v is the velocity.
Since the Ferris wheel is rotating in a vertical circle, the passenger's velocity will be tangential to the circle at any given instant. The magnitude of the velocity can be found using the formula:
v = r * ω
Where v is the velocity, r is the radius of the Ferris wheel, and ω is the angular velocity.
The angular velocity can be calculated using the formula:
ω = 2π / T
Where ω is the angular velocity, and T is the time taken for one revolution.
Given that the radius is 42.2 m and the time for one revolution is 33.2 seconds, we can substitute these values into the equation to calculate the angular velocity:
ω = 2π / 33.2
ω ≈ 0.18996 rad/s
Now we can substitute the values of the acceleration (-2.2864 m/s^2) and velocity (42.2 * 0.18996) into the formula for the angle:
θ = arccos(-2.2864 / (42.2 * 0.18996))
Using a calculator, we find that the angle θ is approximately 95.65 degrees.
Therefore, the angle between the passenger's velocity vector and acceleration vector at that instant is approximately 95.65 degrees.