Final answer:
The angular acceleration of a gramophone record that goes from rest to 60 rev/min in 4 seconds is π/2 rad/s².
Step-by-step explanation:
The student is asked to find the angular acceleration of a gramophone record that accelerates from rest to 60 revolutions per minute (rev/min) in 4.0 seconds
. Assuming the angular acceleration is constant, we can use the formula α = Δω / Δt, where α is the angular acceleration, Δω is the change in angular velocity, and Δt is the change in time.
Firstly, we must convert the angular velocity from revolutions per minute (rev/min) to radians per second (rad/s).
Since 1 revolution is 2π radians and there are 60 seconds in a minute, we can calculate the final angular velocity (ω) as:
Since the record starts from rest, the initial angular velocity (ω0) is 0 rad/s.
Therefore, the change in angular velocity (Δω) is simply the final velocity, which is 2π rad/s.
Using the given time interval (Δt = 4.0 s), we find the angular acceleration (α) to be:
α = Δω / Δt
= 2π rad/s / 4.0 s
= π/2 rad/s²
The angular acceleration of the gramophone record is π/2 rad/s².