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boyakko [2]
3 years ago
12

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

s 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= 11 times while accelerating uniformly.

t1 = 11.9 seconds
n2 = 11 times

Part (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?
Physics
1 answer:
Degger [83]3 years ago
4 0

Answer:

a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

d)\alpha_2=4.1454\,rad.s^{-2}

e)t_2=1.061\,s

Explanation:

Given that:

  • initial speed of turntable, N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

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