The old-fashioned record player had two speeds: one for long-playing albums (LP) and one for singles. The turntable speed for LPs was 33 1/3 rpm (revolutions per minute). This was equivalent to 100 revolutions every three minutes. The speed of a single was 45 rpm. Suppose it took 4.0 seconds for the turntable to ‘speed up’ from 33 1/3 rpm to 45 rpm when a switch was pressed to change the speed. What would be this angular acceleration in rad/sec2?

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aristocles aristocles · Answer 1 · 2019-10-02T08:12:55+02:00

Answer:

[tex]\alpha = 0.305 rad/s^2[/tex]

Explanation:

initial frequency of revolution is given as

[tex]f_1 = 100/3 rpm[/tex]

now initial angular speed is

[tex]\omega_i = 2\pi f[/tex]

[tex]\omega_i = 2\pi(\frac{100}{3\times 60})[/tex]

[tex]\omega_i = 3.49 rad/s[/tex]

Similarly final angular speed is given as

[tex]\omega_f = 2\pi f_2[/tex]

[tex]\omega_f = 2\pi(\frac{45}{60})[/tex]

[tex]\omega_f = 4.71 rad/s[/tex]

Now angular acceleration is given as

[tex]\alpha = \frac{\omega_f - \omega_i}{\Delta t}[/tex]

[tex]\alpha = \frac{4.71 - 3.49}{4}[/tex]

[tex]\alpha = 0.305 rad/s^2[/tex]