A spinning wheel steadily slows from an initial angular velocity of 1.65 rev/s to 0.485 rev/s in 12.2 s.
(a) Calculate the wheel's angular acceleration in radians per second squared.
rad/s2
(b) What angle does it go through during that time?
rad
To answer these questions, we can use the equations of rotational motion.
(a) The angular acceleration (α) can be calculated using the formula:
α = (ωf - ωi) / t
where ωf and ωi are the final and initial angular velocities, and t is the time taken.
ωf = 0.485 rev/s
ωi = 1.65 rev/s
t = 12.2 s
First, we need to convert the angular velocities from revolutions per second (rev/s) to radians per second (rad/s).
1 revolution = 2π radians
Therefore, we have:
ωf = 0.485 rev/s * 2π rad/rev = 0.485 * 2π rad/s
ωi = 1.65 rev/s * 2π rad/rev = 1.65 * 2π rad/s
Now we can substitute these values into the formula:
α = (0.485 * 2π rad/s - 1.65 * 2π rad/s) / 12.2 s
Simplifying the expression, we get:
α = (-1.165 * 2π) rad/s / 12.2 s
α ≈ -0.382 rad/s²
Therefore, the wheel's angular acceleration is approximately -0.382 rad/s².
(b) To calculate the angle (θ) the wheel goes through during the given time, we can use the formula:
θ = ωi * t + (1/2) * α * t^2
ωi = 1.65 rev/s * 2π rad/rev = 1.65 * 2π rad/s
t = 12.2 s
α = -0.382 rad/s²
Substituting these values into the formula:
θ = (1.65 * 2π rad/s) * 12.2 s + (1/2) * (-0.382 rad/s²) * (12.2 s)^2
Simplifying the expression, we get:
θ ≈ 127.437 rad
Therefore, the wheel goes through approximately 127.437 radians during the given time.
Vo=1.65 rev/s * 6.28rad/rev=10.362 rad/s
V=0.485rev/s * 6.28rad/rev = 3.05 rad/s
a. a = (V-Vo)/t