A phonograph record has an initial angular speed of 33 rev/min . The record slows to 11 rev/min in 2.0 seconds. What is the records angular acceleration in rad/s2 during this time interval ?

Respuesta :

In order to calculate the angular acceleration, we can use the following formula:

[tex]a=\frac{v_f-v_i}{t}[/tex]

Where vf is the final angular speed, vi is the initial angular speed and t is the interval of time.

Since the speed is in rev/min, we need to convert to rad/s.

Knowing that 1 rev = 2π rad and 1 min = 60 s, we have:

[tex]\begin{gathered} 33\text{ rev/min}=33\cdot\frac{2\pi\text{ rad}}{60\text{ s}}=3.456\text{ rad/s} \\ 11\text{ rev/min}=11\cdot\frac{2\pi\text{ rad}}{60\text{ s}}=1.152\text{ rad/s} \end{gathered}[/tex]

Now, using vf = 1.152, vi = 3.456 and t = 2, we have:

[tex]a=\frac{1.152-3.456}{2}=\frac{-2.304}{2}=-1.152\text{ rad/s2}[/tex]

So the angular acceleration is -1.152 rad/s².