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Ulleksa [173]
4 years ago
6

A cooling fan is turned off when it is running at 850 rev/min. It turns 1500 revolutions before it comes to a stop. (a) What was

the fan's angular acceleration, assumed constant? (b) How long did it take the fan to come to a complete stop?
Physics
1 answer:
8_murik_8 [283]4 years ago
5 0

Answer

given,

cooling fan revolution = 850 rev/min

fan turns before revolution = 1500 revolutions

\omega = 850 \dfrac{2\pi}{60}

\omega = 89\ rad/s

θ = 1500 revolution

θ = 1500 x 2 x π

θ = 9424.78 rad

a) using equation of rotation

ω² = ω₀² + 2 α θ

ω = 0 because body comes to rest

0 = 89² + 2 x α x 9424.78

α = -0.42 rad/s²

b) time take for the fan to stop

ω = ω₀ + α t

0 = 89 - 0.42 t

t = \dfrac{89}{0.42}

t = 211.9 s

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(a) 1.43 m/s

We can solve this problem by using the law of conservation of energy.

The initial total energy stored in the spring-mass system is

E=U=\frac{1}{2}kx^2

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x=5.08 cm = 0.0508 m

Substituting,

E=\frac{1}{2}(7.91)(0.0508)^2=0.0102 J

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W_f = -(0.0323)(0.145)=-4.68\cdot 10^{-3} J

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(b) +5.08 cm

The speed of the ball is maximum at the instant when all the elastic potential energy stored in the spring has been released: in fact, after that moment, the spring does no longer release any more energy, so the kinetic energy of the ball from that moment will start to decrease, due to the effect of the work done by friction.

The elastic potential energy of the spring is

U=\frac{1}{2}kx^2

And this has all been released when it becomes zero, so when x = 0 (equilibrium position of the spring). However, the spring was initially compressed by 5.08 cm, so the ball has maximum speed when

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The maximum speed is the speed of the ball at the moment when the kinetic energy is maximum, i.e. when all the elastic potential energy has been released.

As we calculated in part (a), the total energy released by the spring is

E = 0.0102 J

The work done by friction here is just the work done to cover the distance of

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Therefore

W_f = -(0.0323)(0.0508)=-1.64\cdot 10^{-3} J

So, the kinetic energy of the ball at the point of maximum speed is

K_f = E+W_f = 0.0102 - 1.64\cdot 10^{-3}=0.00856 J

And so the final speed is

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00856)}{0.00538}}=1.78 m/s

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