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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

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.

One force acting on a machine part is \(\overrightarrow{F} = (-5.00 N)\hat{\imath} + (4.00 N)\hat{\jmath}\). The vector from the origin to the point where the force is applied is \(\overrightarrow{r} = (-0.450 m)\hat{\imath} + (0.150 m)\hat{\jmath}\). (a) In a sketch,show \(\overrightarrow{r}, \overrightarrow{F},\) and the origin. (b) Use the right- hand rule to determine the direction of the torque. (c) Calculate the vector torque for an axis at the origin produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth's mass \(M\), for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg \(\cdot\) m\(^2\) about a vertical axle through its center, and it turns with negligible friction. (a) A child applies an 18.0-N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest, what is its angular speed after this 15.0-s interval? (b) How much work did the child do on the merry-go-round? (c) What is the average power supplied by the child?

A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axle exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

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