Connect with knowledgeable individuals and find the best answers at IDNLearn.com. Our experts provide accurate and detailed responses to help you navigate any topic or issue with confidence.

A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at t = 0,
(a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and in revolutions.
(b) What is the angular speed of the wheel at t = 2.00 s?
(c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles?


Sagot :

Answer:

a) θ = 11 rad, θ = 1.75 rev., b)  w = 9 rad / s, c)  θ = 7.17 rev

Explanation:

This is a rotation kinematics exercise

          θ = θ₀ + w₀ t + ½ α t²

They indicate the initial angular velocity w₀ = 2.00 rad / s, the angular acceleration α = 3.50 rad / s² and that at the initial instant θ₀ = 0

a) let's find the rotated angle

         θ = 0 + 2.00 2.00 +1/2 3.5 2²

         θ = 11 rad

let's reduce 2π rad = 1 rev

        θ = 11 rad (1 rev / 2π rad)

        θ = 1.75 rev.

b) angular velocity

          w = w₀ + α t

          w = 2.00 + 3.50 2

          w = 9 rad / s

c) the angular displacement to reach this speed

          w² = w₀² + 2 α θ

         

in this case they indicate that w = 2  9 = 18 rad / s

          θ = [tex]\frac{w^2 - w_o^2}{2 \alpha }[/tex]

          θ = [tex]\frac{18^2 - 2^2 }{2 \ 3.5 }[/tex]

          θ = 45.7 rad

let's reduce to rev

          θ = 45.7 rad (1rev / 2π rad)

          θ = 7.17 rev

Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.