Question #56595

QUESTION NO. 5 &6 on
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Expert's answer

2016-03-09T09:04:05-0500

Answer on Question #56595, Physics Mechanics Relativity

The flywheel of a large motor in a factory has mass 30kg30\,\mathrm{kg} and moment of inertia 67.5kgm267.5\,\mathrm{kg}\cdot \mathrm{m}^2 about rotation axis. The motor develops a constant torque of 600Ntm600\,\mathrm{Nt}\cdot \mathrm{m}, and the flywheel starts from rest.

(a) What is the angular acceleration of the flywheel?

(b) What is its angular velocity after making 4 revolutions?

(c) How much work is done by the motor during the first 4 revolutions?

(d) What is the average power output of the motor during the first 4 revolutions?

Solution

(a)


M=JεM = J \varepsilon


where M=600NmM = 600\,\mathrm{N}\cdot \mathrm{m} is a constant torque; ε\varepsilon is the angular acceleration of the flywheel; J=67.5kgm2J = 67.5\,\mathrm{kg}\cdot \mathrm{m}^2 is moment of inertia about rotation axis.

Then


ε=M/J=600Nm/67.5kgm28.89rad/s2\varepsilon = M / J = 600\,\mathrm{N}\cdot \mathrm{m} / 67.5\,\mathrm{kg}\cdot \mathrm{m}^2 \approx 8.89\,\mathrm{rad} / \mathrm{s}^2


(b)

Angular velocity:


ω=ω0+εt\omega = \omega_0 + \varepsilon tω0=0 (flywheel starts from rest).\omega_ {0} = 0 \text{ (flywheel starts from rest)}.


The angle of rotation


φ=φ0+ω0t+12εt2\varphi = \varphi_ {0} + \omega_ {0} t + \frac {1}{2} \varepsilon t ^ {2}


where ω0=0\omega_0 = 0 and φ0=0\varphi_0 = 0 (flywheel starts from rest).

4 revolutions φ=42π\varphi = 4\cdot 2\pi, then from (3)


t=16π/ε=4π/ε=2.38st = \sqrt {16 \pi / \varepsilon} = 4 \sqrt {\pi / \varepsilon} = 2.38s


Angular velocity after making 4 revolutions:


ω=εt=ε4π/ε=4πε=4π8.8921.14rad/s\omega = \varepsilon t = \varepsilon 4 \sqrt {\pi / \varepsilon} = 4 \sqrt {\pi \varepsilon} = 4 \sqrt {\pi \cdot 8.89} \approx 21.14rad / s


(c)

The work is done by the motor during the first 4 revolutions:


A=Jω22Jω022=Jω220=Jω22=67.58.8922=2667.33J=2.67kJA = \frac {J \omega^ {2}}{2} - \frac {J \omega_ {0} ^ {2}}{2} = \frac {J \omega^ {2}}{2} - 0 = \frac {J \omega^ {2}}{2} = \frac {67.5 \cdot 8.89 ^ {2}}{2} = 2667.33J = 2.67kJ


(d)

The average power output of the motor during the first 4 revolutions:


P=At=2667.332.38=1120Watt\left\langle P \right\rangle = \frac {A}{t} = \frac {2667.33}{2.38} = 1120\text{Watt}


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