Question

A DVD is initially at rest so that the line PQ on the disks surface is along the +x-axis. The disc begins to turn with a constant LaTeX: \alpha_zα z= 5.0 rad/s2. At t = 0.40 s, what is the angle between the line PQ and the +x-axis

Answer 1

The angle between the line PQ on a DVD's surface (initially aligned with the +x-axis) and the +x-axis after 0.40 seconds, with a constant angular acceleration of 5.0 rad/s^2, is 0.4 radians.

Step-by-step explanation:

The question involves calculating the angle between the line PQ on the DVD's surface (which is initially along the +x-axis) and the +x-axis at a later time, given a constant angular acceleration. To find the angle at a specific time t, we use the equation for angular displacement θ, which is θ = ω_0 × t + 0.5 × α × t^2, where θ is the angular displacement, ω_0 is the initial angular velocity (which is zero in this case), α is the angular acceleration, and t is the time elapsed. With an angular acceleration (α) of 5.0 rad/s^2 and a time (t) of 0.40s, we can calculate the angle. Since the DVD starts from rest, the initial angular velocity ω_0 is 0.

The formula simplifies to θ = 0.5 × α × t^2. Substituting in the values we have: θ = 0.5 × 5.0 rad/s^2 × (0.40 s)^2 = 0.5 × 5.0 × 0.16 = 0.4 rad. This is the angle between the line PQ and the +x-axis after 0.40 seconds have passed.