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Chapter 8: Problem 26

A Ferris wheel rotates because a motor exerts a torque on the wheel. The radius of the London Eye, a huge observation wheel on the banks of the Thames, is \(67.5 \mathrm{m}\) and its mass is \(1.90 \times 10^{6} \mathrm{kg} .\) The cruising angular speed of the wheel is $3.50 \times 10^{-3} \mathrm{rad} / \mathrm{s} .$ (a) How much work does the motor need to do to bring the stationary wheel up to cruising speed? [Hint: Treat the wheel as a hoop.] (b) What is the torque (assumed constant) the motor needs to provide to the wheel if it takes 20.0 s to reach the cruising angular speed?

Short Answer

Expert verified
Answer: The motor needs to do approximately \(1.8\times 10^7 ~\mathrm{J}\) of work to bring the stationary wheel up to cruising speed, and it needs to exert a torque of about \(1.5\times 10^6 ~\mathrm{N\cdot m}\) to reach the cruising angular speed in 20 seconds.

Step by step solution

01

Calculate the Moment of Inertia for the Wheel (Hoop)

For a hoop, the moment of inertia (I) can be calculated using the formula: \[I = M R^2\] where M is the mass of the hoop and R is its radius. Given, the mass M = \(1.9 \times 10^6 ~\mathrm{kg}\) and radius R = 67.5 m. I = (\(1.9 \times 10^6 ~\mathrm{kg}\))(\(67.5 ~\mathrm{m}\))^2 Calculating this gives us I ≈ \(8.6 \times 10^9 ~\mathrm{kg\cdot m^2}\).
02

Calculate the Initial and Final Rotational Kinetic Energies

The work-energy theorem states that the work done on a system is equal to the change in its kinetic energy. In the case of a rotating object, we can write the work-energy theorem as: Work = \( \frac{1}{2} I\omega_f^2 - \frac{1}{2} I\omega_i^2\) Since the wheel is initially stationary, the initial angular velocity (\(\omega_i\)) is zero, so the equation simplifies to : Work = \(\frac{1}{2}I\omega_f^2\) Where I is the moment of inertia (\(8.6 \times 10^9 ~\mathrm{kg\cdot m^2}\)) and \(\omega_f\) is the final angular velocity (\(3.50 \times 10^{-3} ~\mathrm{rad/s}\)).
03

Calculate the Work Done by the Motor

Plugging in the values for I and \(\omega_f\) into the equation above and solving for the work: Work = \(\frac{1}{2}\times (8.6 \times 10^9 ~\mathrm{kg\cdot m^2})\times (3.50 \times 10^{-3} ~\mathrm{rad/s})^2\) Calculating this gives us Work ≈ \(1.8 \times 10^{7} ~\mathrm{J}\).
04

Calculate the Angular Acceleration

To find the angular acceleration (α), we can use the formula: \(\omega_f = \omega_i + \alpha t\) Since the wheel is initially stationary, the initial angular velocity (\(\omega_i\)) is zero. Given, \(\omega_f = 3.50 \times 10^{-3} ~\mathrm{rad/s}\) and time t = 20 s. Solving for α: α = \(\frac{(3.50 \times 10^{-3} ~\mathrm{rad/s})}{20 ~\mathrm{s}}\) Calculating this gives us α ≈ \(1.75 \times 10^{-4} ~\mathrm{rad/s^2}\).
05

Calculate the Torque

Now we can calculate the torque (τ) exerted by the motor using the formula: \(\tau = I \alpha\) where I is the moment of inertia (\(8.6 \times 10^9 ~\mathrm{kg\cdot m^2}\)) and α is the angular acceleration (\(1.75 \times 10^{-4} ~\mathrm{rad/s^2}\)). \(\tau = (8.6 \times 10^9 ~\mathrm{kg\cdot m^2}) (1.75 \times 10^{-4} ~\mathrm{rad/s^2})\) Calculating this gives us τ ≈ \(1.5 \times 10^6 ~\mathrm{N\cdot m}\). The motor needs to do (a) approximately \(1.8\times 10^7 ~\mathrm{J}\) of work to bring the stationary wheel up to cruising speed, and (b) needs to exert a torque of about \(1.5\times 10^6 ~\mathrm{N\cdot m}\) to reach the cruising angular speed in 20 seconds.

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