Question

Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get up to its full speed of f₁ = 78 revolutions per minute. Case 2: The DJ then changes the speed of the turntable from f₁ = 78 to f₂ = 120 revolutions per minute. She notices that the turntable rotates exactly n₂= 15 times while accelerating uniformly.t₁ = 11.9 secondsn₂ = 11 timesPart (a) Calculate the angular speed of the turntable while it is turning at f₁ = 78 in radians/second in Case 1.Part (b) How many revolutions does the turntable make while accelerating in Case 1?Part (c) Calculate the magnitude of the angular acceleration of the turntable in Case 1, in radians/second².Part (d) Calculate the magnitude of the angular acceleration of the turntable (in radians/second²) while increasing to 120 RPM (Case 2).Part (e) How long (in seconds) does it take for the turntable to go from f₁ = 78 to f₂ = 120 RPM?

Answer 1

Answer:

Part a)

$\omega = 8.17 rad/s$

Part b)

$N = 7.74 rev$

Part c)

$\alpha = 0.69 rad/s^2$

Part d)

$\alpha = 0.48 rad/s^2$

Part e)

$t = 9.14 s$

Explanation:

Part a)

Angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi(\frac{78}{60})$

$\omega = 8.17 rad/s$

Part b)

Since turn table is accelerating uniformly

so we will have

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{8.17 + 0}{2}(11.9)$

$2N\pi = 48.6$

$N = 7.74 rev$

Part c)

angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{8.17 - 0}{11.9}$

$\alpha = 0.69 rad/s^2$

Part d)

When its angular speed changes to 120 rpm

then we will have

$\omega_2 = 2\pi (\frac{120}{60})$

$\omega_2 = 12.56 rad/s$

number of turns revolved is 15 times

so we have

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

$12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)$

$\alpha = 0.48 rad/s^2$

Part e)

now for uniform acceleration we have

$\omega_f - \omega_i = \alpha t$

$12.56 - 8.17 = 0.48 t$

$t = 9.14 s$