Problem 37 - University Physics with Modern Physics

Question

A uniform bar has two small balls glued to its ends. The bar is 2.00 \(\mathrm{m}\) long and has mass 4.00 \(\mathrm{kg}\) , while the balls each have mass 0.500 \(\mathrm{kg}\) and can be treated as point masses. Find the moment of 0.500 \(\mathrm{kg}\) and can be treated as point masses. Find the moment of inertia of this combination about each of the following axes: (a) an axis perpendicular to the bar through its center; (b) an axis perpendicularto the bar through one of the balls; (c) an axis parallel to the bar through both balls; (d) an axis parallel to the bar and 0.500 \(\mathrm{m}\) m from it.

Step-by-Step Solution

Answer

(a) 2.33 kg·m², (b) 7.33 kg·m², (c) 1.00 kg·m², (d) 1.25 kg·m².

Step 1: Find the Moment of Inertia for Axis A

For the axis perpendicular to the bar through its center:The moment of inertia of a bar about its center is given by:\[ I_{bar} = \frac{1}{12} m L^2 \]where \( m = 4.00 \ \mathrm{kg} \) and \( L = 2.00 \ \mathrm{m} \).Plugging in the values:\[ I_{bar} = \frac{1}{12} \times 4.00 \ \mathrm{kg} \times (2.00 \ \mathrm{m})^2 = \frac{1}{12} \times 4 \times 4 = \frac{16}{12} = 1.33 \ \mathrm{kg \cdot m^2} \]The moment of inertia of each ball with respect to the center is:\[ I_{ball} = m r^2 \]where \( r = 1.00 \ \mathrm{m} \) (since each ball is at the end of the bar).Therefore:\[ I_{ball, total} = 2 \times 0.500 \ \mathrm{kg} \times (1.00 \ \mathrm{m})^2 = 1.00 \ \mathrm{kg \cdot m^2} \]Thus, the total moment of inertia is:\[ I_{total} = I_{bar} + I_{ball, total} = 1.33 + 1.00 = 2.33 \ \mathrm{kg \cdot m^2} \]

Step 2: Find the Moment of Inertia for Axis B

For the axis perpendicular to the bar through one of the balls:The moment of inertia of the bar about this axis using the parallel axis theorem is:\[ I' = I_{bar} + md^2 \]where \( d = 1.00 \ \mathrm{m} \) (distance from the center to the axis).Thus:\[ I' = \frac{1}{12} \times 4.00 \ \mathrm{kg} \times (2.00 \ \mathrm{m})^2 + 4.00 \times (1.00 \ \mathrm{m})^2 = 1.33 + 4.00 = 5.33 \ \mathrm{kg \cdot m^2} \]Only one ball contributes directly as it rotating at the axis, so:\[ I_{ball} = 0.500 \ \mathrm{kg} \times (0 \ \mathrm{m})^2 = 0 \]The remaining ball's contribution:\[ I_{ball, distant} = 0.500 \ \mathrm{kg} \times (2.00 \ \mathrm{m})^2 = 2.00 \ \mathrm{kg \cdot m^2} \]Thus, the total moment of inertia is:\[ I_{total} = I' + I_{ball, distant} = 5.33 + 2.00 = 7.33 \ \mathrm{kg \cdot m^2} \]

Step 3: Find the Moment of Inertia for Axis C

For the axis parallel to the bar through both balls:The entire mass of the system rotates around the axis at a distance of the bar's length from the axis.For the bar:\[ I_{bar} = 4.00 \ \mathrm{kg} \times (0 \ \mathrm{m})^2 = 0 \]The balls, however, contribute:\[ I_{ball,total} = 2 \times 0.500 \ \mathrm{kg} \times (1.00 \ \mathrm{m})^2 = 1.00 \ \mathrm{kg \cdot m^2} \]Thus, the total moment of inertia is:\[ I_{total} = I_{bar} + I_{ball,total} = 0 + 1.00 = 1.00 \ \mathrm{kg \cdot m^2} \]

Step 4: Find the Moment of Inertia for Axis D

For the axis parallel to the bar and 0.500 m from it:The entire mass of the system is away from the axis.Both the bar and balls contribute at a distance of 0.500 m:For the bar:\[ I_{bar} = 4.00 \ \mathrm{kg} \times (0.500 \ \mathrm{m})^2 = 1.00 \ \mathrm{kg \cdot m^2} \]Each ball contributes:\[ I_{balls} = 2 \times 0.500 \ \mathrm{kg} \times (0.500 \ \mathrm{m})^2 = 0.25 \ \mathrm{kg \cdot m^2} \]Thus, the total moment of inertia is:\[ I_{total} = I_{bar} + I_{balls} = 1.00 + 0.25 = 1.25 \ \mathrm{kg \cdot m^2} \]

Key Concepts

Parallel Axis Theorem

The Parallel Axis Theorem is a fundamental concept in calculating moments of inertia for complex systems. It allows us to calculate the moment of inertia of an object about an axis that is parallel to an axis through its center of mass.

The theorem states:

• If you know the moment of inertia (\(I_c\)) about an axis through the center of mass, you can find the moment of inertia (\(I\)) about any parallel axis a distance \(d\) away.
  \[ I = I_c + md^2 \]
  
• Here, \(m\) is the mass of the object and \(d\) is the perpendicular distance between the two axes.

The Parallel Axis Theorem is particularly useful for rotated or translated systems where direct calculation would be complex. It's essential for engineers and scientists working with rotational dynamics, as it simplifies calculations when transferring known data to practical setups.

Point Masses

In rotational dynamics, a point mass is an idealized object with mass concentrated at a single point, simplifying calculations of inertia.

When calculating the moment of inertia of systems with point masses, each mass contributes to the total inertia:

• The contribution of each point mass \(m\) is calculated by \( I = mr^2 \), where \(r\) is the distance from the rotation axis.
  For example, in the case of the bar end balls, each ball on either end can be treated as a point mass.
  
• This method aids in simplifying the problem, turning complex-shaped bodies into collections of point masses.

This approximation is invaluable in physics, especially when dealing with spherical masses, particles, or even small rigid bodies, as it simplifies the mathematics without significant accuracy loss.

Rotational Dynamics

Rotational dynamics is the branch of physics that deals with the motion of rotating bodies and the forces and torques that affect them.

Key concepts include:

• Torque (\(\tau\)): The force that causes rotational motion, calculated by \(\tau = rF \sin \theta\). It's akin to force in linear dynamics.
• Angular momentum (\(L\)): The rotational equivalent of linear momentum, defined as \(L = I \omega\), where \(\omega\) is the angular velocity.
• Moment of Inertia (\(I\)): The resistance to change in rotational motion, equivalent to mass in linear motion.

Understanding rotational dynamics helps in predicting how an object will spin, accelerate, or maintain its motion under various conditions such as gravity or other external forces.

In practical examples like spinning bars or wheels, these concepts allow for the design and analysis of rotating systems, crucial in engineering and daily life applications.