Question

A 41.0-cm diameter disk rotates with a constant angular acceleration of 2.90 rad/s². It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this time.

(a) At t = 2.49 s, find the angular speed of the wheel.

(b) At t = 2.49 s, find the magnitude of the linear velocity AND tangential acceleration of P.

(c) At t = 2.49 s, find the position of P (in degrees, with respect to the positive x-axis).

Answer 1

As the disk rotates, its angular velocity increases at the rate of 2.9 rad/s each second. To determine its angular velocity at 2.49 seconds, use the following equation.  

ωf = ωi + α * t, ωi = 0
ωf = 2.9 * 2.49 = 7.22 rad/s
Since point P is on the rim of the disk, the distance from the center to this point is the radius of the disk.

r = 0.5 * 0.41 = 0.205 meter
To convert angular velocity to linear velocity, multiply by this number.  

v = 7.22 * 0.205 = 1.4801 m/s

Its linear velocity increased from 0 m/s to 1.4801 in 2.49 seconds. To determine its tangential acceleration, divide its linear velocity by the time.  

a = 1.4801 ÷ 2.49 = 0.594418 m/s^2

Use the following equation to determine the angle the point has rotated in 2.49 seconds.  

θ = ½ * (ωi + ωf) * t, ωi = 0
θ = ½ * 7.22 * 2.49 = 8.99 radians = 8.99 x 180˚/π = 515.35⁰

To determine the angle respect to the positive x-axis, subtract 360˚

θ = 515.35⁰ - 360⁰ = 155.35⁰

The total angle is 155.35⁰ + 57.3⁰ = 212.65⁰ Answer