Angular kinematics

Question

A wheel 2.20 m in diameter lies in a vertical plane and rotates with a constant angular acceleration of 4.10 rad/s². The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s find the following.

(a) the angular speed of the wheel in rad/s

(b) the tangential speed of the point P in m/s

Answer 1

For a diameter of 2.20 m, the radius (r) is 1.10 m. α = 4.10 rad/s², θ_i = 57.3° = 1.00 radians, ω_i = 0 rad/s

a) Given Δt = 2, find ω_f.

Use the kinematic equation for constant angular acceleration: ω_f = ω_i + αt

ω_f = ω_i + αt

ω_f = 0 + (4.10)(2) = 8.20 rad/s

b) Use v_T = rω

v_T = rω

v_T = (1.10)(8.20) = 9.02 m/s