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Chapter 10: Problem 68

A \(4.00-\mathrm{kg}\) mass and a \(3.00-\mathrm{kg}\) mass are attached to opposite ends of a thin 42.0 -cm-long horizontal rod (Fig. \(10-60\) ). The system is rotating at angular speed \(\omega=5.60 \mathrm{rad} / \mathrm{s}\) about a vertical axle at the center of the rod. Determine \((a)\) the kinetic energy \(K\) of the system, and (b) the net force on each mass. (c) Repeat parts \((a)\) and \((b)\) assuming that the axle passes through the \(\mathrm{CM}\) of the system.

Short Answer

Expert verified
(a) Calculate the kinetic energy using inertia: \( K = \frac{1}{2} I \omega^2 \). (b) The net force is centripetal: \( F = m \omega^2 r \). For CM, recalculate inertia and force.

Step by step solution

01

Determine System Moment of Inertia

To find the kinetic energy, start by calculating the moment of inertia of the system about the given axis (center of the rod). The masses are point particles, so the moment of inertia is computed as \( I = m_1r_1^2 + m_2r_2^2 \), where \( m_1 = 4.00 \text{ kg} \), \( m_2 = 3.00 \text{ kg} \), and both \( r_1 \) and \( r_2 \) are 21.0 cm (half of 42.0 cm), converted to meters (0.21 m). Substitute these values to get \[ I = (4.00 \, \text{kg})(0.21 \, \text{m})^2 + (3.00 \, \text{kg})(0.21 \, \text{m})^2 \].
02

Calculate Initial Kinetic Energy

Using the moment of inertia from Step 1, calculate the kinetic energy \( K \) of the system using the formula \( K = \frac{1}{2}I\omega^2 \), where \( \omega = 5.60 \, \text{rad/s} \). Substitute \( I \) and \( \omega \) into the equation to find the kinetic energy.
03

Compute Centripetal Force on Each Mass

The net force on each mass is the centripetal force, given by \( F = m\omega^2r \). For each mass (\( m_1 = 4.00 \, \text{kg} \) and \( m_2 = 3.00 \, \text{kg} \)), compute \( F_1 \) and \( F_2 \) using their respective radii (0.21 m) and the angular speed \( \omega = 5.60 \, \text{rad/s} \).
04

Shift the Reference Axis to the Center of Mass (CM)

To find the moment of inertia about the CM, first calculate the center of mass location using \( x_{CM} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} \). Recalculate \( r_1 \) and \( r_2 \) for each mass relative to the CM. Then determine the new moment of inertia \( I_{CM} = m_1(r_1')^2 + m_2(r_2')^2 \).
05

Calculate Kinetic Energy About CM

Using the new moment of inertia about the CM from Step 4, recalculate the kinetic energy using \( K_{CM} = \frac{1}{2}I_{CM}\omega^2 \).
06

Calculate Net Force on Each Mass About CM

Use the new radii calculated from the CM position to find the net force \( F_{CM} = m\omega^2r_{CM} \) for each mass individually.

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Key Concepts

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

Moment of Inertia
The moment of inertia is a key concept in rotational dynamics. It acts as rotational mass, showing how an object's mass is distributed with respect to its axis of rotation. For point masses, the moment of inertia can be calculated using the formula: \[ I = \sum{m_i r_i^2} \]Here's what each symbol means:
  • \( I \) is the moment of inertia,
  • \( m_i \) is the mass of each point particle, and
  • \( r_i \) is the distance from the axis of rotation.
In situations where masses are attached to a rod, like in the exercise, it is important to calculate the distance from the axis of rotation accurately. Using this understanding, we calculate the system's moment of inertia about the given axis by summing over each mass effectively. This helps us investigate how the arrangement of masses impacts rotational motion.
Kinetic Energy in Rotational Systems
Kinetic energy in rotational motion is analogous to kinetic energy in linear motion. However, for rotating objects, the formula adapts to account for rotation:\[ K = \frac{1}{2} I \omega^2 \]Let's break it down:
  • \( K \) is the kinetic energy of the system,
  • \( I \) is the moment of inertia, and
  • \( \omega \) is the angular velocity.
In rotational systems, kinetic energy depends both on how mass is spread out from the axis (moment of inertia) and the angular speed. As seen in the exercise, once we have determined the moment of inertia, computing kinetic energy becomes straightforward. Multiplying the moment of inertia by the square of the angular velocity gives us the rotational kinetic energy for this system.
Centripetal Force Calculation
Centripetal force is the net force required to keep an object moving in a circular path. This force points towards the center of the circle.The formula for centripetal force is:\[ F = m \omega^2 r \]Here's what each part represents:
  • \( F \) is the centripetal force,
  • \( m \) is the mass of the object,
  • \( \omega \) is the angular velocity, and
  • \( r \) is the radius of the circular path.
In the given exercise, each mass experiences a centripetal force due to its rotation around the central axis. By substituting the values for mass, angular speed, and the radius, the centripetal force is determined, ensuring each mass remains on its circular path around the axis.
Center of Mass
The center of mass (CM) is a point representing the average position of the entire mass of a system. For ease of analysis, especially with multiple objects, it is crucial to determine the center of mass.For two point masses, the center of mass can be calculated using:\[ x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]In this formula:
  • \( x_{CM} \) is the center of mass position,
  • \( m_1, m_2 \) are the masses, and
  • \( x_1, x_2 \) are the positions of these masses.
In scenarios like the exercise, once we determine the CM, we can recalculate the moment of inertia with respect to this new axis through the CM. This shift provides a different value for the kinetic energy. It also helps us understand force distribution and motion when the system's rotation changes.

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Most popular questions from this chapter

(I) Use the parallel-axis theorem to show that the moment of inertia of a thin rod about an axis perpendicular to the rod at one end is \(I=\frac{1}{3} M \ell^{2},\) given that if the axis passes through the center, \(I=\frac{1}{12} M \ell^{2}(\) Fig. 20 \(\mathrm{f}\) and \(\mathrm{g})\)

An oxygen molecule consists of two oxygen atoms whose total mass is \(5.3 \times 10^{-26} \mathrm{~kg}\) and whose moment of inertia about an axis perpendicular to the line joining the two atoms, midway between them, is \(1.9 \times 10^{-46} \mathrm{~kg} \cdot \mathrm{m}^{2}\). From these data, estimate the effective distance between the atoms.

(a) A grinding wheel \(0.35 \mathrm{~m}\) in diameter rotates at 2500 rpm. Calculate its angular velocity in rad/s. (b) What are the linear speed and acceleration of a point on the edge of the grinding wheel?

Pilots can be tested for the stresses of flying high-speed jets in a whirling "human centrifuge," which takes 1.0 min to turn through 20 complete revolutions before reaching its final speed. ( \(a\) ) What was its angular acceleration (assumed constant), and ( \(b\) ) what was its final angular speed in rpm?

Figure 65 illustrates an \(\mathrm{H}_{2} \mathrm{O}\) molecule. The \(\mathrm{O}-\mathrm{H}\) bond length is 0.096 \(\mathrm{nm}\) and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) bonds make an angle of \(104^{\circ} .\) Calculate the moment of inertia for the \(\mathrm{H}_{2} \mathrm{O}\) mole cule about an axis passing through the center of the oxygen atom (a) perpendicular to the plane of the molecule, and (b) in the plane of the molecule, bisecting the H- \(\mathrm{O}-\mathrm{H}\) bonds.
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