A Ferris wheel, rotating initially at an angular speed of 0.500 rad/s, accelerates over a 7.00-s interval at a rate of 0.04 rad/s^2. What angular displacement does the Ferris wheel undergo in this 7-s interval?
To find the angular displacement of the Ferris wheel, we can use the formula:
θ = ωi * t + (1/2) * α * t^2
Where:
θ is the angular displacement,
ωi is the initial angular speed,
t is the time interval, and
α is the angular acceleration.
Given:
ωi = 0.500 rad/s
t = 7.00 s
α = 0.04 rad/s^2
Let's substitute these values into the formula:
θ = (0.500 rad/s) * (7.00 s) + (1/2) * (0.04 rad/s^2) * (7.00 s)^2
θ = 3.500 rad + (1/2) * 0.04 rad/s^2 * 49.00 s^2
θ = 3.500 rad + 0.98 rad
θ = 4.480 rad
Therefore, the Ferris wheel undergoes an angular displacement of 4.480 radians in this 7-second interval.
To calculate the angular displacement of the Ferris wheel, we can use the formula:
θ = ω0t + (1/2)αt^2
Where:
θ is the angular displacement
ω0 is the initial angular velocity
α is the angular acceleration
t is the time interval
Given:
ω0 = 0.500 rad/s (initial angular speed)
α = 0.04 rad/s^2 (angular acceleration)
t = 7.00 s (time interval)
Let's substitute the given values into the formula and calculate the angular displacement:
θ = (0.500 rad/s)(7.00 s) + (1/2)(0.04 rad/s^2)(7.00 s)^2
First, calculate the first term:
(0.500 rad/s)(7.00 s) = 3.50 rad
Next, calculate the second term:
(1/2)(0.04 rad/s^2)(7.00 s)^2 = 0.98 rad
Now, add the two terms together to find the angular displacement:
θ = 3.50 rad + 0.98 rad = 4.48 rad
Therefore, the angular displacement of the Ferris wheel during the 7-second interval is 4.48 radians.