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Q6.14:
To maintain a rotor at a uniform angular speed of 200 rad (s^{-1}), an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.

Solution :

Step 1: Identification of Given Variables

Based on the problem statement, we define the following physical quantities:

  • Angular velocity ($\omega$): $200 \text{ rad s}^{-1}$
  • Applied torque ($\tau$): $180 \text{ N m}$

Step 2: Theoretical Foundation

The power ($P$) delivered by a torque ($\tau$) to a system rotating at an angular velocity ($\omega$) is defined by the rotational analogue of the linear power equation ($P = F \cdot v$). [Per the definition of rotational power in rigid body dynamics]:

$P = \tau \cdot \omega$

Step 3: Dimensional and Scalar Calculation

Substituting the given values into the equation:

$P = 180 \text{ N m} \times 200 \text{ rad s}^{-1}$

$P = 36000 \text{ N m rad s}^{-1}$

Since a radian is a dimensionless quantity ($1 \text{ rad} = 1$), and the product of Newton-meters per second equates to Joules per second, the unit reduces to Watts ($\text{W}$):

$P = 36000 \text{ W}$

Step 4: Conversion to Standard Metric Prefixes

To express the result in a more standard engineering format, we convert the value to kilowatts ($\text{kW}$):

$P = \frac{36000 \text{ W}}{1000} = 36 \text{ kW}$

Vector Representation of Rotational Power

ω = 200 rad/s τ = 180 N·m Power Calculation: P = τ × ω Result: 36,000 W (36 kW)

Final Solution: The power required by the engine to maintain the uniform angular speed is $36,000 \text{ W}$ or $36 \text{ kW}$.


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