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A table-top fan has radius of 0.5 m. It starts to rotate from rest to 800 rpm within 30 seconds. Determine: 1) the angular momentum 2) how many revolutions have it gone through after 2 minutes? 3) If the rotational inertia of each blade around the center is 0.3 kg.m^ 2, what is the magnitude of the torque provided by the motor of the fan, assuming no friction?

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Answer:

Part a)

L = 25.13 kg m^2/s

Part b)

N = 3200 rev

Part c)

torque = 0.837 Nm

Step-by-step explanation:

Part a)

As we know that angular frequency is given as


\omega = 2\pi f


\omega = 2\pi(800/60)


\omega = 83.77 rad/s

now the angular momentum is given as


L = I\omega


L = (0.3)(83.77)


L = 25.13 kg m^2/s

Part b)

angular acceleration of fan is given as


\alpha = 2\pi(800/60)/(30)


\alpha = 2.79 rad/s^2

total number of revolutions are


N = (1)/(4\pi)\alpha t^2


N = (1)/(4\pi)(2.79)(120^2)


N = 3200 rev

Part c)

Torque is given as


\tau = I\alpha


\tau = (0.30)(2.79) = 0.837 Nm

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