Question

A small turntable is initially at rest. It then obtains an angular acceleration of 4 rad/s2 for 3 s. After that, its angular acceleration is zero for the next 10 s. Find the change in its angular position during that entire time period.

Answer 1

The total change in the angular position of the turntable over the entire 13 second period is 138 radians.

Step-by-step explanation:

To calculate the change in the angular position during the entire time period, we need to consider two segments: when the turntable is accelerating and when it is rotating with a constant angular velocity.

Acceleration Phase (0 to 3s)

During the first 3 seconds, the turntable has an angular acceleration (α) of 4 rad/s². The change in angular position (ΔΘ) in this phase can be calculated using the kinematic equation ΔΘ = ω0t + 0.5αt², where ω0 is the initial angular velocity, t is the time, and α is the angular acceleration. Since it starts from rest, ω0 = 0, so the equation simplifies to ΔΘ = 0.5 × 4 rad/s² × (3s)² = 18 rad.

Constant Speed Phase (3s to 13s)

The angular velocity at the end of the acceleration phase becomes the constant angular velocity for the next 10 seconds. This final angular velocity (ω) can be calculated by ω = ω0 + αt = 0 + 4 rad/s² × 3s = 12 rad/s. The change in angular position for the constant speed phase is ΔΘ = ωt = 12 rad/s × 10s = 120 rad.

The total change in angular position over the entire 13 seconds is the sum of the two phases: ΔΘtotal = ΔΘacceleration phase + ΔΘconstant speed phase = 18 rad + 120 rad = 138 rad.