At t = 0, a wheel rotating about a fixed axis at a constant angular deceleration of 0.37 rad/s2 has an angular velocity of 0.5 rad/s and an angular position of 8.5 rad.
What is the angular position of the wheel
after 2 s?
Answer in units of rad
To find the angular position of the wheel after 2 seconds, we can use the following kinematic equation:
θ = θ0 + ω0t + (1/2)αt^2
Where:
θ = final angular position
θ0 = initial angular position
ω0 = initial angular velocity
α = angular acceleration
t = time
Given:
θ0 = 8.5 rad (initial angular position)
ω0 = 0.5 rad/s (initial angular velocity)
α = -0.37 rad/s^2 (angular deceleration)
t = 2 s (time)
Substituting the given values into the equation, we have:
θ = 8.5 + (0.5)(2) + (1/2)(-0.37)(2)^2
θ = 8.5 + 1 - 0.37(2)^2
θ = 8.5 + 1 - 0.37(4)
θ = 8.5 + 1 - 1.48
θ = 8.5 - 0.48
θ ≈ 8.02 rad
Therefore, the angular position of the wheel after 2 seconds is approximately 8.02 rad.
To find the angular position of the wheel after 2 seconds, we can use the equation of angular motion:
θ = θ₀ + ω₀t + (1/2)αt²
Where:
θ is the final angular position
θ₀ is the initial angular position (given as 8.5 rad)
ω₀ is the initial angular velocity (given as 0.5 rad/s)
t is the time (given as 2 s)
α is the angular deceleration (given as -0.37 rad/s² since it's deceleration)
Now, let's substitute the given values into the equation and solve for θ:
θ = 8.5 rad + (0.5 rad/s)(2 s) + (0.5)(-0.37 rad/s²)(2 s)²
= 8.5 rad + 1 rad - 0.614 rad
= 8.886 rad
Therefore, the angular position of the wheel after 2 seconds is approximately 8.886 radians.
angle (in radians) = 8.5 + 0.5*t - (0.37/2)*t^2
Substitute t = 2 and solve.
A minus sign is on the last term because the wheel is decelerating.