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If the body's center of mass were not placed on the rotational axis of the turntable, how would the person's measured moment of inertia compare to the moment of inertia for rotation about the center of mass? (a) The measured moment of inertia would be too large; (b) the measured moment of inertia would be too small; (c) the two moments of inertia would be the same; (d) it depends on where the body's center of mass is placed relative to the center of the turntable.

Short Answer

Expert verified
The measured moment of inertia would be too large (option a).

Step by step solution

01

Understand the Concept of Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotation. It depends on the mass distribution relative to the axis of rotation. If mass is farther from the axis, the moment of inertia increases.
02

Apply the Parallel Axis Theorem

The parallel axis theorem states that the moment of inertia about any axis parallel to one through the center of mass is given by: \( I = I_{cm} + Md^2 \), where \( I_{cm} \) is the moment of inertia through the center of mass, \( M \) is the mass, and \( d \) is the distance between the axes.
03

Analyze the Problem

In this problem, the body's center of mass is away from the rotational axis of the turntable. According to the parallel axis theorem, the moment of inertia measured around the turntable's axis will be \( I = I_{cm} + Md^2 \), which is greater than \( I_{cm} \) since \( Md^2 > 0 \).
04

Conclusion

The measured moment of inertia is greater when the axis of rotation is not through the center of mass. Therefore, the correct answer is (a) The measured moment of inertia would be too large.

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Key Concepts

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

Center of Mass
The center of mass (COM) of an object is a critical point where the entire mass seems to be concentrated. It is the average location of all the mass in a body. Imagine it as the balancing point of an object where it would stay level if supported.
This concept is crucial in understanding rotational movements and balance.
  • If a force is applied at the center of mass, the object will move without rotating.
  • If the force is applied elsewhere, it will cause the object to rotate around the center of mass.
Understanding the center of mass is essential when discussing the moment of inertia, as it affects how easily an object can be spun around an axis.
Parallel Axis Theorem
The parallel axis theorem is a fundamental concept in rotational dynamics. It provides a way to calculate an object's moment of inertia when the rotational axis is not through the center of mass. According to this theorem:
  • The moment of inertia about an axis parallel to one through the center of mass can be calculated with the equation: \[ I = I_{cm} + Md^2 \]
  • \( I_{cm} \) is the moment of inertia through the center of mass.
  • \( M \) is the total mass of the object.
  • \( d \) is the distance between the two axes.
This theorem shows that the moment of inertia increases when the axis is farther from the center of mass. Hence, the farther the rotational axis is from the center of mass, the larger the contribution of \( Md^2 \) to the total moment of inertia.
Rotational Axis
The rotational axis is the line around which an object rotates. The position of this axis plays a significant role in the rotational dynamics of the body.
  • It can pass through the center of mass, leading to minimal resistance to rotation.
  • Or it can be offset, causing increased resistance due to the need for more force to rotate the object.
Choosing the right axis of rotation is crucial in engineering and design, as it affects the effort needed to spin or rotate objects. If the axis does not pass through the center of mass, the object will experience a larger moment of inertia, making it more resistant to changes in its rotational motion.
Mass Distribution
Mass distribution refers to how mass is spread throughout an object. It has a direct impact on the moment of inertia. The further the mass is from the rotational axis, the greater the moment of inertia will be.
  • For a concentrated mass closer to the rotational axis, the moment of inertia is lower.
  • For a spread-out mass farther from the axis, the moment of inertia increases significantly.
This idea is essential in understanding why the moment of inertia can change with different mass arrangements. Engineers and designers must consider mass distribution in their work, especially when designing rotating systems like wheels or turbines.

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Most popular questions from this chapter

A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 m and a total mass of 120 kg. The turntable is initially rotating at 3.00 rad/s about a vertical axis through its center. Suddenly, a 70.0-kg parachutist makes a soft landing on the turntable at a point near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal?

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_x\) and \(a_z\) are approximately zero and \(v_x\) and \(\omega_z\) are approximately constant. Rolling without slipping means \(v_x = r\omega_z\) and \(a_x = r\alpha_z\) . If an object is set in motion on a surface \(without\) these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\), rotating with angular speed \(\omega_0\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_k\). (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_x\) of the center of mass and \(a_z\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_z = \omega_0\) but \(v_x =\) 0. Rolling without slipping sets in when \(v_x = r\omega_z\) . Calculate the \(distance\) the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

You are designing a slide for a water park. In a sitting position, park guests slide a vertical distance \(h\) down the waterslide, which has negligible friction. When they reach the bottom of the slide, they grab a handle at the bottom end of a 6.00-m-long uniform pole. The pole hangs vertically, initially at rest. The upper end of the pole is pivoted about a stationary, frictionless axle. The pole with a person hanging on the end swings up through an angle of 72.0\(^\circ\), and then the person lets go of the pole and drops into a pool of water. Treat the person as a point mass. The pole's moment of inertia is given by \(I = \frac{1}{3} ML^2\), where \(L =\) 6.00 m is the length of the pole and \(M =\) 24.0 kg is its mass. For a person of mass 70.0 kg, what must be the height h in order for the pole to have a maximum angle of swing of 72.0\(^\circ\) after the collision?

The solid wood door of a gymnasium is 1.00 m wide and 2.00 m high, has total mass 35.0 kg, and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 N to the door for 8.00 ms. Find the angular speed of the door after the impact. [\(Hint:\) Integrating Eq. (10.29) yields \(\Delta L_z = \int_{t1}^{t2} (\Sigma\tau_z)dt = (\Sigma\tau_z)_av \Delta t\). The quantity \(\int_{t1}^{t2} (\Sigma\tau_z)dt\) is called the angular impulse.]

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes \(higher\) with friction on the right side than without friction?

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