Question

A ceiling fan has four blades, each with a mass of \(0.35 \mathrm{kg}\) and a length of \(60 \mathrm{cm} .\) Model each blade as a rod connected to the fan axle at one end. When the fan is turned on, it takes 4.35 s for the fan to reach its final angular speed of 1.8 rev/s. What torque was applied to the fan by the motor? Ignore torque due to the air.

Short answer

Answer: The torque applied to the fan by the motor is approximately \(0.437\,\text{N m}\).

Step-by-step solution

Calculate the moment of inertia of one fan blade

Model each blade as a rod connected to the fan axle at one end. The moment of inertia of a rod with mass \(m\) and length \(L\) rotating about one end is given by the formula: \(I = \frac{1}{3}mL^2\). Given the mass \(m = 0.35\,\text{kg}\) and length \(L = 60\,\text{cm}\) (in meters, \(L = 0.6\,\text{m}\)), we can find the moment of inertia for one blade: \(I = \frac{1}{3}(0.35\,\text{kg})(0.6\,\text{m})^2 \approx 0.042\,\text{kg m}^2\)

Calculate the total moment of inertia of the fan

The fan has 4 blades, so the total moment of inertia \(I_{\text{total}}\) is given by the sum of the moments of inertia of each individual blade: \(I_{\text{total}} = 4I = 4(0.042\,\text{kg m}^2) = 0.168\,\text{kg m}^2\)

Find the angular acceleration of the fan

The fan starts at an angular speed of \(0\,\text{rev/s}\) and takes \(4.35\,\text{s}\) to reach its final angular speed of \(1.8\,\text{rev/s}\). Recall that in the SI unit, angular speed is typically expressed in radians/second. To convert from rev/s to rad/s, we use the conversion factor \(\text{1 rev} = 2\pi\,\text{rad}\): \(\omega = 1.8\,\text{rev/s} \times \frac{2\pi\,\text{rad}}{\text{1 rev}} \approx 11.31\,\text{rad/s}\) Now we can find the angular acceleration \(\alpha\) using the equation \(\alpha = \frac{\Delta\omega}{\Delta t}\). In this case, \(\Delta\omega=\omega\) and \(\Delta t = 4.35\,\text{s}\): \(\alpha = \frac{11.31\,\text{rad/s}}{4.35\,\text{s}} \approx 2.60\,\text{rad/s}^2\)

Determine the torque applied by the motor

Finally, we can determine the torque \(\tau\) applied by the motor using the relation between torque, moment of inertia, and angular acceleration: \(\tau = I_{\text{total}}\alpha\): \(\tau = (0.168\,\text{kg m}^2)(2.60\,\text{rad/s}^2) \approx 0.437\,\text{N m}\) Therefore, the torque applied to the fan by the motor is approximately \(0.437\,\text{N m}\).