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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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Most popular questions from this chapter

Derive the rotational form of Newton's second law as follows. Consider a rigid object that consists of a large number \(N\) of particles. Let \(F_{i}, m_{i},\) and \(r_{i}\) represent the tangential component of the net force acting on the ith particle, the mass of that particle, and the particle's distance from the axis of rotation, respectively. (a) Use Newton's second law to find \(a_{i}\), the particle's tangential acceleration. (b) Find the torque acting on this particle. (c) Replace \(a_{i}\) with an equivalent expression in terms of the angular acceleration \(\alpha\) (d) Sum the torques due to all the particles and show that $$\sum_{i=1}^{N} \tau_{i}=I \alpha$$
An experimental flywheel, used to store energy and replace an automobile engine, is a solid disk of mass \(200.0 \mathrm{kg}\) and radius $0.40 \mathrm{m} .\( (a) What is its rotational inertia? (b) When driving at \)22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph}),$ the fully energized flywheel is rotating at an angular speed of \(3160 \mathrm{rad} / \mathrm{s} .\) What is the initial rotational kinetic energy of the flywheel? (c) If the total mass of the car is \(1000.0 \mathrm{kg},\) find the ratio of the initial rotational kinetic energy of the flywheel to the translational kinetic energy of the car. (d) If the force of air resistance on the car is \(670.0 \mathrm{N},\) how far can the car travel at a speed of $22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph})$ with the initial stored energy? Ignore losses of mechanical energy due to means other than air resistance.
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.
A merry-go-round (radius \(R\), rotational inertia \(I_{\mathrm{i}}\) ) spins with negligible friction. Its initial angular velocity is \(\omega_{i}\) A child (mass \(m\) ) on the merry-go-round moves from the center out to the rim. (a) Calculate the angular velocity after the child moves out to the rim. (b) Calculate the rotational kinetic energy and angular momentum of the system (merry-go-round + child) before and after.
A flywheel of mass 182 kg has an effective radius of \(0.62 \mathrm{m}\) (assume the mass is concentrated along a circumference located at the effective radius of the flywheel). (a) How much work is done to bring this wheel from rest to a speed of 120 rpm in a time interval of 30.0 s? (b) What is the applied torque on the flywheel (assumed constant)?
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