A wheel of moment of inertia 0⋅10 kg-m2 is rotating about a shaft at an angular speed of 160 rev/minute. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with a common angular speed of 200 rev/minute. Find the moment of inertia of the second wheel.
Answers
moment of inertia of second wheel is 0.4 kgm²
moment of inertia of first wheel, I1 = 0.1 kgm²
angular speed of first wheel, ω1 = 160 rev/min.
angular speed of 2nd wheel, ω2 = 300 rev/min
after coupling of wheels, common angular speed of it , ω = 200 rev/min
as here it is clear that there is no external torque acting on system of wheels. so, angular momentum will be conserved.
i.e., initial angular momentum = final angular momentum
⇒ I1ω1 + I2ω2 = (I1 + I2)ω
⇒0.1 kgm² × 160 rev/min + I2 × 300 rev/min = (0.1 + I2) × 200 rev/min
⇒16 + 300I2 = 20 + 200I2
⇒100I2 = 20 - 16 = 4
⇒I2 = 0.04 kgm²
hence, moment of inertia of second
wheel is 0.4 kgm²
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Answer:
Given
For the first wheel,
I1 = 10 kg-m2 and ω1 = 160 rev/min
For the second wheel, ω2 = 300 rev/min
Let I2 be the moment of inertia of the second wheel.
After they are coupled, we have
ω = 200 rev/min
If we take the two wheels to be an isolated system, we get
Total external torque = 0
Therefore, we have
\[I_1 \omega_1 + I_2 \omega_2 = \left( I_1 + I_2 \right) \omega\]
\[\Rightarrow 0 . 10 \times 160 + I_2 \times 300 = \left( 0 . 10 + I_2 \right) \times 200\]
\[ \Rightarrow 16 + 300 I_2 = 20 + 200 I_2 \]
\[ \Rightarrow 100 I_2 = 4\]
\[ \Rightarrow I_2 = \frac{4}{100} = 0 . 04 kg - m^2\]