Problem 63 - University Physics with Modern Physics

Question

A thin, uniform, \(3.80-\mathrm{kg}\) bar, 80.0 \(\mathrm{cm}\) long, has very small \(2.50-\mathrm{kg}\) balls glued on at either end (Fig. P10.63). It is supported horizontally by a thin, horizontal, frictionless axle passing through its center and perpendicular to the bar. Suddenly the right-hand ball becomes detached and falls off, but the other ball remains glued to the bar. (a) Find the angular acceleration of the bar just after the ball falls off. (b) Will the angular acceleration remain constant as the bar continues to swing? If not, will it increase or decrease? (c) Find the angular velocity of the bar just as it swings through its vertical position.

Step-by-Step Solution

Answer

(a) \(\alpha = 18.17 \; \text{rad/s}^2\); (b) It will decrease; (c) \(\omega = 8.58 \; \text{rad/s}\).

Step 1: Understand the Problem

We have a bar with a mass, two attached balls at each end, and an axle in the middle. One ball detaches, and we need to find how this changes the system's motion. Use rotational dynamics, starting with angular acceleration right after the ball detaches.

Step 2: Define Known Variables and Equations

Mass of bar, \(m_b = 3.80 \text{ kg}\); Mass of each ball, \(m_{ball} = 2.50 \text{ kg}\); Length of bar, \(L = 0.80 \text{ m}\). The moment of inertia of the system prior to the ball falling off is \(I_i = \frac{1}{12}m_bL^2 + 2m_{ball}\left(\frac{L}{2}\right)^2\). The effective moment after the ball falls is \(I_f = \frac{1}{12}m_bL^2 + m_{ball}\left(\frac{L}{2}\right)^2\). Torque is given by \(\tau = I_f \cdot \alpha\), where \(\alpha\) is angular acceleration.

Step 3: Calculate Initial Moment of Inertia

Using the moment of inertia formula for composite systems,\[I_i = \frac{1}{12} \cdot 3.80 \; \text{kg} \cdot (0.80 \; \text{m})^2 + 2 \cdot 2.50 \; \text{kg} \cdot (0.40 \; \text{m})^2\] Calculate to get \(I_i = 1.04 \; \text{kg} \cdot \text{m}^2\).

Step 4: Calculate Final Moment of Inertia

Now calculate the moment of inertia after one ball falls:\[I_f = \frac{1}{12} \cdot 3.80 \; \text{kg} \cdot (0.80 \; \text{m})^2 + 2.50 \; \text{kg} \cdot (0.40 \; \text{m})^2\] This simplifies and calculates to \(I_f = 0.54 \; \text{kg} \cdot \text{m}^2\).

Step 5: Determine the Torque Acting on the System

The torque is only due to the remaining ball at the end. It equals \(\tau = m_{ball} \cdot g \cdot \left(\frac{L}{2}\right)\), where \(g\) is the acceleration due to gravity (9.81 m/s²). Substitute the values to find \(\tau = 2.50 \cdot 9.81 \cdot 0.40 = 9.81 \; \text{Nm}\).

Step 6: Solve for Angular Acceleration (Part A)

Using \(\tau = I_f \cdot \alpha\), solve for \(\alpha\):\[\alpha = \frac{\tau}{I_f} = \frac{9.81}{0.54} \approx 18.17 \; \text{rad/s}^2\].

Step 7: Consider Time-dependent Factors for Angular Acceleration (Part B)

The angular acceleration depends on the angle as it swings down (since \(\tau = m_{ball}g\cdot\frac{L}{2}\cdot \sin\theta\)). It will change as the bar swings, likely decreasing, because torque decreases as the angle goes to 0.

Step 8: Calculate Final Angular Velocity (Part C)

Use conservation of energy: initial potential energy equals the kinetic energy at the vertical bottom. Gravitational potential energy is \((m_{ball}g\cdot\frac{L}{2})\). Kinetic energy is \(\frac{1}{2}I_f\omega^2\).\[m_{ball}g\frac{L}{2} = \frac{1}{2}I_f\omega^2\] Solve for \(\omega\):\[\omega = \sqrt{\frac{2m_{ball}g\cdot0.40}{0.54}} \approx 8.58 \; \text{rad/s}\].

Key Concepts

Angular Acceleration

Angular acceleration is a measure of how quickly an object's rotational speed changes. It is denoted by \( \alpha \) and is expressed in radians per second squared (rad/s²). In our system, when the right-hand ball detaches from the bar, the balance of the system changes, affecting how it spins.

To understand it simply, think of angular acceleration as the rotational equivalent of linear acceleration. When the ball detaches, the bar starts rotating because of an unbalanced force acting as torque. We calculate angular acceleration using the formula:

• \( \alpha = \frac{\tau}{I_f} \)

Here, \( \tau \) is the torque, and \( I_f \) is the moment of inertia after the ball falls. By inserting the values, \( \alpha \) comes out to be approximately 18.17 rad/s².

This means that right after the ball detaches, the bar begins to rotate more rapidly, gaining angular speed quickly. However, as it continues to rotate, this acceleration might change as we'll explore next.

Moment of Inertia

The moment of inertia, often symbolized as \( I \), represents an object's resistance to changes in its rotational motion. It plays a similar role in rotational dynamics as mass does in linear dynamics. For any rotational system, determining the moment of inertia is crucial since it influences the system's angular acceleration.

For our exercise, before any balls fall off, the moment of inertia \( I_i \) accounts for both balls attached to the bar:

• \( I_i = \frac{1}{12}m_bL^2 + 2m_{ball}\left(\frac{L}{2}\right)^2 \)

Using this formula, we find the initial moment of inertia to be 1.04 kg·m². After one ball falls, the formula modifies to:

• \( I_f = \frac{1}{12}m_bL^2 + m_{ball}\left(\frac{L}{2}\right)^2 \)

This yields 0.54 kg·m². Notice the substantial decrease in the moment of inertia once a ball detaches, making it easier for the bar to spin faster due to a lower resistance to rotational change.

Torque

Torque is the rotational equivalent of force, symbolized by \( \tau \). It causes an object to rotate around an axis. In this problem, torque is generated by the weight of the remaining ball pushing down at a distance from the axle.

Torque is calculated using:

• \( \tau = m_{ball} \cdot g \cdot \frac{L}{2} \)

where \( g \) is the acceleration due to gravity (9.81 m/s²). With our values, the torque right after the ball falls computes to 9.81 Nm.

This torque is responsible for initiating the bar's rotational movement. However, as the bar swings and changes its position, the effective torque alters (primarily because the distance change affects the angle to gravity, depicted as \( \sin\theta \) in changing over time). Thus, torque—and thereby angular acceleration—is not constant during the swing. Initially large, it decreases as the bar approaches vertical alignment, making the rotational speed peak and then stabilize.