The applied torque increases the angular speed by the application of an
angular acceleration.
- The speed after 10 seconds is approximately 37.16 rev/s.
Reasons:
The speed of the flywheel at the axis = 4 rev/s
The torque applied, T = 25 N·m
The time the torque is applied, t = 10 sec
Moment of inertia of the flywheel, I = 1.2 kg·m²
Required:
The speed at the end of the 10 seconds
Solution:
T = I·α
Where;
α = Angular acceleration
[tex]\displaystyle \alpha = \frac{T}{I}[/tex]
Therefore;
[tex]\displaystyle \alpha = \frac{24 \ N\cdot m}{1.2 \ kg \cdot m^2} = \mathbf{20\frac{5}{6} \ s^{-2}}[/tex]
The rotational speed, ω = ω₀ + α·t
Which gives;
[tex]\displaystyle \mathrm{The \ angular \ speed, } \ \omega = \frac{2 \cdot \pi \times 4 \ rad }{s} = \frac{8 \cdot \pi \ rad }{s}[/tex]
ω₀ = 8·π rad/s
Which gives;
[tex]\displaystyle \omega = \mathbf{8 \cdot \pi +2 \frac{5}{6} \times 10} = 233.47[/tex]
The speed of the wheel in revolution per second is therefore;
[tex]\displaystyle Speed \ in \ rev/s = \frac{8 \cdot \pi +2 \frac{5}{6} \times 10}{2\cdot \pi} \approx 37.16[/tex]
The speed after 10 seconds is approximately 37.16 rev/s.
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