Question

Initially a wheel rotating about a fixed axis at a constant angular deceleration of 0.5 rad/s 2 has an angular velocity of 0 rad/s and an angular position of 6.1 rad. What is the angular position of the wheel after 3.9 s

Answer 1

Answer:

$\theta = 5.83\ rad$

Step-by-step explanation:

given,

angular deceleration, α = -0.5 rad/s²

final angular velocity,ω_f = 0 rad/s

angular position, θ = 6.1 rad

angular position at 3.9 s = ?

now, Calculating the initial angular speed

$\omega_f^2 = \omega_i^2 + 2 \alpha \theta$

$0 = \omega_i^2 - 2\times 0.5\times 6.1$

$\omega_i = \sqrt{6.1}$

$\omega_i = 2.47\ rad/s$

now, angular position calculation at t=3.9 s

$\theta = \omega_i t + \dfrac{1}{2}\alpha t^2$

$\theta =2.47\times 3.9 - \dfrac{1}{2}\times 0.5\times 3.9^2$

$\theta = 5.83\ rad$

Hence, the angular position of the wheel after 3.9 s is equal to 5.83 rad.