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Chapter 10: Problem 64

(II) A rotating uniform cylindrical platform of mass 220 \(\mathrm{kg}\) and radius 5.5 \(\mathrm{m}\) slows down from 3.8 rev/s to rest in 16 \(\mathrm{s}\) when the driving motor is disconnected. Estimate the power output of the motor (hp) required to maintain a steady speed of 3.8 \(\mathrm{rev} / \mathrm{s}\) .

Short Answer

Expert verified
The required power output of the motor is approximately 79.3 hp.

Step by step solution

01

Understanding the Initial Condition

First, note that the platform is initially rotating at 3.8 revolutions per second. We need to convert this to angular velocity in radians per second (rad/s). Since there are \(2\pi\) radians in one revolution, substitute to get \(\omega_i = 3.8 \times 2\pi\, \text{rad/s}\).
02

Calculating Initial Angular Velocity

Calculate the initial angular velocity using:\[ \omega_i = 3.8 \times 2\pi = 23.876 \text{ rad/s} \]
03

Calculating Moment of Inertia

For a uniform cylindrical platform, the moment of inertia \( I \) is computed as:\[ I = \frac{1}{2} m r^2 \] where \( m = 220 \text{ kg} \) and \( r = 5.5 \text{ m} \), obtaining:\[ I = \frac{1}{2} \times 220 \times 5.5^2 = 3327.5 \text{ kg} \cdot \text{m}^2 \]
04

Calculating Initial Rotational Kinetic Energy

The initial kinetic energy of the rotating platform is given by:\[ KE_i = \frac{1}{2} I \omega_i^2 \]Substitute values:\[ KE_i = \frac{1}{2} \times 3327.5 \times 23.876^2 = 946736.8618 \text{ J} \]
05

Finding Power Output Required

Since the motor must supply energy equivalent to the kinetic energy to maintain the speed, the power output is \( P = \frac{KE_i}{t} \), where \( t = 16 \text{ s} \) is the time for the motor to deliver this energy:\[ P = \frac{946736.8618}{16} = 59171.054 \text{ W} \]
06

Converting Power to Horsepower

Finally, convert the power from watts to horsepower. Recall that 1 horsepower (hp) is equivalent to 746 watts:\[ P_{hp} = \frac{59171.054}{746} \approx 79.3 \text{ hp} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Velocity
Angular velocity helps us understand how fast an object is rotating. It's like speed, but for rotation.
In the exercise, we start with a platform that rotates at 3.8 revolutions per second (rev/s).
To work with angular velocity, we need to switch from revolutions per second to radians per second, since
  • One revolution is equal to \(2\pi\) radians.
  • We multiply by \(2\pi\) to convert: \(\omega_i = 3.8 \times 2\pi = 23.876\, \text{rad/s}\).
This gives us the initial angular speed \(\omega_i\) in radians per second, allowing for more accurate and useful calculations in physics.
Moment of Inertia
The moment of inertia is like mass in linear motion. It tells us how hard it is to change the rotation speed.
It depends on both the mass and the shape of the object.For a cylinder, like our platform, the moment of inertia \( I \) is calculated using the formula:
  • \( I = \frac{1}{2} m r^2 \)
  • where \( m = 220 \text{ kg} \) and \( r = 5.5 \text{ m} \).
  • This results in \( I = 3327.5 \text{ kg} \cdot \text{m}^2 \).
This quantity shows how resistant the platform is to changes in its rotational motion.
Rotational Kinetic Energy
When an object spins, it has kinetic energy due to its rotation.
This is known as rotational kinetic energy, similar to how moving objects have translational kinetic energy.The formula for rotational kinetic energy \( KE \) is:
  • \( KE = \frac{1}{2} I \omega^2 \)
  • where \( I = 3327.5 \text{ kg} \cdot \text{m}^2 \) and \( \omega_i = 23.876 \text{ rad/s} \).
Plugging the values into the equation gives us the initial rotational kinetic energy:
  • \( KE_i = \frac{1}{2} \times 3327.5 \times 23.876^2 = 946736.8618 \text{ J} \).
This energy tells us how much effort the system is doing to keep spinning at its current speed.
Power Conversion
Power is the rate at which energy is transferred or converted.
In our case, the power output of the motor is needed to sustain the platform鈥檚 rotation at a steady speed.To find the power required, we consider the energy lost due to slowing down. This energy equals the initial rotational kinetic energy that needs to be supplied constantly:
  • Power, \( P = \frac{KE_i}{t} = \frac{946736.8618}{16} = 59171.054 \text{ W} \)
But, often power is expressed in horsepower for mechanical contexts:
  • Since 1 horsepower (hp) is 746 watts, we convert: \( P_{hp} = \frac{59171.054}{746} \approx 79.3 \text{ hp} \).
This tells us how much power the motor needs to provide to keep the platform spinning without slowing down.

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Most popular questions from this chapter

(II) A uniform thin rod of length \(\ell\) and mass \(M\) is suspended freely from one end. It is pulled to the side an angle \(\theta\) and released. If friction can be ignored, what is angular velocity, and the speed of its free end, at the lowest point?

(III) An Atwood's machine consists of two masses, \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}}\) which are connected by a massless inelastic cord that passes over a pulley, Fig. \(57 .\) If the pulley has radius \(R\) and moment of inertia \(I\) about its axle, determine the acceleration of the masses \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}},\) and compare to the situation in which the moment of inertia of the pulley is ignored. [Hint: The tensions \(F_{\mathrm{TA}}\) and \(F_{\mathrm{TB}}\) are not equal. With the Atwood machine, assume \(I=0\) for the pulley.]

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