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An engineer wants to find the rotational inertia of an oddshaped object of mass \(11.3 \mathrm{~kg}\) about an axis through its center of mass. The object is supported with a wire through its center of mass and along the desired axis. The wire has a torsional constant \(\kappa=0.513 \mathrm{~N} \cdot \mathrm{m} .\) The engineer observes that this torsional pendulum oscillates through \(20.0\) cycles in \(48.7 \mathrm{~s}\). What value of the rotational inertia is calculated?

Short Answer

Expert verified
The calculated value for the rotational inertia is \(0.1586 kg \cdot m^2\)

Step by step solution

01

Calculate Period of Oscillation

The first step is to calculate the period of single oscillation. The time for 20 cycles is given as 48.7s. To find the time for a single oscillation, we divide the total time by the number of cycles: \(T = \frac{48.7s}{20} = 2.435s\)
02

Derive equation for I

The next step is to rearrange the formula for the period of oscillation to solve for I (moment of inertia). \(T = 2\pi\sqrt{\frac{I}{\kappa}}\) Squaring both sides, we get \(T^2 = (2\pi)^2\frac{I}{\kappa}\). Rearranging to solve for I gives us \(I = \frac{T^2\kappa}{(2\pi)^2}\)
03

Substitute values

Finally, substituting the value of \(T = 2.435s\) and \(\kappa = 0.513N \cdot m\) into the equation from step 2, we get \(I = \frac{(2.435s)^2 * 0.513N \cdot m}{(2\pi)^2} = 0.1586 kg \cdot m^2\)

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Key Concepts

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

Torsional Pendulum
Imagine a swinging door that is not opening and closing, but rather twisting back and forth about its hinges. This is similar to how a torsional pendulum behaves. A torsional pendulum oscillates due to a wire or a rod's twisting motion when an object attached to it rotates.

In our exercise, the engineer uses a torsional pendulum to find the rotational inertia of an odd-shaped object. By observing how the object rotates around its center of mass and is supported by a wire, which twists as it oscillates, the engineer can gather crucial data to measure rotational inertia. Using the properties of the torsional pendulum, such as the relationship between the twisting force and the angle of twist, helps us solve important physical problems like the one in this exercise.

Key points to remember:
  • Torsional pendulums rotate through twisting motion.
  • They are useful for measuring properties like rotational inertia.
Period of Oscillation
The period of oscillation is the time it takes for one complete cycle of oscillation to occur. In this exercise, the period is a crucial component to determining the object's rotational inertia. The engineer observed 20 cycles in 48.7 seconds, so to find the period of one complete oscillation, we simply divide 48.7 seconds by 20. This calculation gives us a period, denoted by the symbol \( T \), of approximately 2.435 seconds.

The formula used here links the period to the torsional pendulum and will help us solve for the object's moment of inertia later. Such relationships are common in physics, allowing one to deduce unknown quantities from measured data. Keeping track of the period is essential for applications in engineering, as it helps to understand dynamic behaviors in systems influenced by oscillatory motion.

Remember to:
  • Calculate the period as total time divided by the number of cycles.
  • Use the period in further calculations for oscillatory systems.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to rotational motion around an axis. It is more than just a category for objects with mass and size; it's a fundamental property in rotational dynamics that affects how an object spins.

In our exercise, we need to calculate the moment of inertia \( I \) of the object using the torsional pendulum's characteristics. By rearranging the formula from step two, we find that:\[ I = \frac{T^2 \kappa}{(2\pi)^2} \]where \( T \) is the period of oscillation and \( \kappa \) is the torsional constant. Substituting in known values yields \( I = 0.1586 \, kg \cdot m^2 \).This number describes the efficiency of energy distribution in rotatory systems.

Key insights:
  • The moment of inertia varies with mass distribution.
  • It's crucial for predicting rotational behavior.
Torsional Constant
The torsional constant, denoted by \( \kappa \), characterizes how much torque is required to twist an object by a certain angle. It plays a significant role in the dynamics of a torsional pendulum, influencing the period of oscillation.

In our specific scenario, the torsional constant is 0.513 N·m, which represents the wire's resistance to twisting. A higher torsional constant means the wire is stiffer and harder to twist, which typically results in a shorter period. Conversely, a smaller \( \kappa \) value indicates more flexibility and a longer period. This relationship is vital for calculating the object's rotational inertia through the oscillatory motion generated by the torsional pendulum.

Essential takeaways:
  • The torsional constant helps determine the ease of twisting.
  • It affects how quickly or slowly a system oscillates.

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Most popular questions from this chapter

Two particles execute simple harmonic motion of the same amplitude and frequency along the same straight line. They pass one another when going in opposite directions each time their displacement is half their amplitude. Find the phase difference between them.

A pendulum whose upper end is attached so as to allow the pendulum to swing freely in any direction can be used to repeat an experiment first shown publicly by Foucault in Paris in 1851 . If the pendulum is set oscillating, the plane of oscillation slowly rotates with respect to a line drawn on the floor, even though the tension in the wire supporting the bob and the gravitational pull of the Earth on the bob lie in a vertical plane. (a) Show that this is a result of the fact that the Earth is not an inertial reference frame. ( \(b\) ) Show that for a Foucault pendulum at a latitude \(\theta\), the period of rotation of the plane, in hours, is \(24 \sin \theta .(c)\) Explain in simple terms the result at \(\theta=90^{\circ}\) (the poles) and \(\theta=0^{\circ}\) (the equator).

An oscillator consists of a block of mass \(512 \mathrm{~g}\) connected to a spring. When set into oscillation with amplitude \(34.7 \mathrm{~cm}\), it is observed to repeat its motion every \(0.484 \mathrm{~s}\). Find \((a)\) the period, \((b)\) the frquency, \((c)\) the angular frequency, \((d)\) the force constant, \((e)\) the maximum speed, and \((f)\) the maximum force exerted on the block.

The scale of a spring balance reading from 0 to \(50.0 \mathrm{lb}\) is \(4.00\) in. long. A package suspended from the balance is found to oscillate vertically with a frequency of \(2.00 \mathrm{~Hz}\). How much does the package weigh?

A \(4.00-\mathrm{kg}\) block is suspended from a spring with a force constant of \(5.00 \mathrm{~N} / \mathrm{cm}\). A \(50.0-\mathrm{g}\) bullet is fired into the block from below with a speed of \(150 \mathrm{~m} / \mathrm{s}\) and comes to rest in the block. (a) Find the amplitude of the resulting simple harmonic motion. (b) What fraction of the original kinetic energy of the bullet appears as mechanical energy in the oscillator?

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