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. \(\mathbf{P 1 0 . 5 7}\) ). 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.

Short answer

(a) Angular acceleration is \(49.26 \ rad/s^2\). (b) It decreases. (c) Angular velocity is \(14.02 \ rad/s\).

Step-by-step solution

Determine Initial Moment of Inertia

The initial moment of inertia of the system includes the bar and both attached balls. The moment of inertia for the bar alone is \( I_{bar} = \frac{1}{12} m L^2 \), where \( m = 3.8 \ kg \) and \( L = 0.8 \ m \). The moment of inertia for the two balls is \( 2 \times m_b r^2 = 2 \times 2.5 \ kg \times (0.4 \ m)^2 \). Therefore, the initial moment of inertia is: \[ I_{initial} = \frac{1}{12} (3.8) (0.8)^2 + 2 \times 2.5 \times (0.4)^2 \]. Calculate this to get \( I_{initial} = 0.5067 \ kg \cdot m^2 \).

Determine Final Moment of Inertia

Once the right-hand ball falls off, we need the moment of inertia of the new system. Now, the moment of inertia of just the bar and one ball is \( I_{final} = \frac{1}{12} (3.8) (0.8)^2 + 2.5 \times (0.4)^2 \). Calculate this to find \( I_{final} = 0.1987 \ kg \cdot m^2 \).

Use Torque and Angular Acceleration Formula

The torque due to gravity on the attached ball is \( \tau = m_b g r = 2.5 \times 9.8 \times 0.4 \). The angular acceleration is given by \( \alpha = \frac{\tau}{I_{final}} \). Plug in \( \tau = 9.8 \ N \cdot m \) and \( I_{final} = 0.1987 \ kg \cdot m^2 \) to find \( \alpha = 49.26 \ rad/s^2 \).

Determine Changes in Angular Acceleration

As the bar swings, the torque due to the gravitational force on the ball will change because the effective lever arm changes. It will reach a maximum when horizontal and decrease to zero when vertical, meaning the angular acceleration will decrease.

Calculate Final Angular Velocity

Use conservation of energy. Initial potential energy is converted to rotational kinetic energy as the bar swings. Set potential energy \( m_b g h \) equal to rotational kinetic energy \( \frac{1}{2} I_{final} \omega^2 \). So, \( 2.5 \times 9.8 \times 0.4 = \frac{1}{2} \times 0.1987 \times \omega^2 \). Solve for \( \omega \) to find \( \omega = \sqrt{\frac{2 \times 2.5 \times 9.8 \times 0.4}{0.1987}} \approx 14.02 \ rad/s \).

Key concepts

Moment of Inertia

The concept of moment of inertia is essential to understanding how objects resist rotational motion. Think of it as the rotational equivalent of mass in linear motion. In simple terms, an object with a large moment of inertia is harder to spin. The moment of inertia depends not only on the mass of the object but also on how this mass is distributed relative to the axis of rotation.

For the given exercise, the initial moment of inertia of the system was calculated including both the bar and the two balls. The formula \[ I_{bar} = \frac{1}{12} m L^2 \] is used to find the moment of inertia of the bar. The term \( m \) represents its mass, and \( L \) is its length. Each attached ball contributes an additional factor to the moment of inertia, given by \[ 2 \times m_b \times r^2 \]where \( m_b \) is the mass of the ball and \( r \) is the distance from the axis.

After one ball detaches, the moment of inertia decreases because there is less mass resisting the change in rotational motion, making it easier for the remaining parts to rotate. The final moment of inertia of the bar and remaining ball is recalculated to adjust for this change.

Rotational Kinetic Energy

Rotational kinetic energy is the energy possessed by a rotating object. It is the counterpart of kinetic energy in linear motion. For rotating objects, this energy is given by the formula:\[ KE_{rotational} = \frac{1}{2} I \omega^2 \]where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

In the context of the exercise, as the bar with one ball loses potential energy falling from the horizontal position, it converts this energy into rotational kinetic energy. Since the ball moves in a circular path due to the force of gravity, the system's increase in velocity is captured through the angular velocity \( \omega \).

Understanding how energy transforms from potential to kinetic in scenarios like these highlights the importance of rotational kinetic energy. Despite the complexity, the core principle is that energy shifts from one form to another during motion, without being lost.

Conservation of Energy

The conservation of energy theorem is one of the most powerful principles in physics. It states that the total energy in a closed system remains constant, though it may transform from one form to another. In rotational dynamics, this principle explains how potential energy can be transformed into rotational kinetic energy.

For this problem, initially, when the bar is held horizontally, the system has gravitational potential energy due to the height of the center of mass from the ground. As the bar swings downward, this potential energy decreases and converts into rotational kinetic energy. The formula used to express this transformation is:\[ m_b g h = \frac{1}{2} I_{final} \omega^2 \]where \( m_b \) is the mass of the ball, \( g \) is the gravitational acceleration, and \( h \) is the height.

By rearranging this equation, you can find the angular velocity \( \omega \) at the lowest point of the swing, knowing it equates potential energy at the top. This showcases a fundamental application of energy conservation, illustrating that despite the motion and change in form, the total energy remains constant.