Question

A bullet of mass \(m_{\mathrm{B}}=1.00 \cdot 10^{-2} \mathrm{~kg}\) is moving with a speed of \(100 . \mathrm{m} / \mathrm{s}\) when it collides with a rod of mass \(m_{\mathrm{R}}=5.00 \mathrm{~kg}\) and length \(L=1.00 \mathrm{~m}\) (shown in the figure). The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass. The bullet embeds itself in the rod at a distance \(L / 4\) from the pivot point. As a result, the bullet-rod system starts rotating. a) Find the angular velocity, \(\omega\), of the bullet-rod system after the collision. You can neglect the width of the rod and can treat the bullet as a point mass. b) How much kinetic energy is lost in the collision?

Short answer

The angular velocity of the bullet-rod system after the collision is approximately 0.60 rad/s. b) How much kinetic energy is lost during the collision? The loss in kinetic energy during the collision is 42.50 J.

Step-by-step solution

Find the initial angular momentum of the bullet

We can find the initial angular momentum of the bullet about the pivot point using the following equation: \( L_1 = m_B \times v \times r \) where \(L_1\) is the initial angular momentum of the bullet, \(m_B\) is the mass of the bullet, \(v\) is the initial speed of the bullet, and \(r=\frac{L}{4}\) is the distance from the pivot point where the bullet gets embedded in the rod. \(L_1 = (1.00 \times 10^{-2} \mathrm{kg}) (100 \mathrm{m/s}) (\frac{1}{4} \mathrm{m}) = 0.25 \mathrm{kg \cdot m^2/s}\)

Apply conservation of angular momentum

The total initial angular momentum equals the total final angular momentum. The rod's initial angular momentum is zero, so the total initial angular momentum is equal to the bullet's initial angular momentum: \(L_{1} = L_{2}\) where \(L_2\) is the final angular momentum of the bullet-rod system.

Find the final angular momentum of the bullet-rod system

We can write the formula for the final angular momentum of the bullet-rod system as: \(L_{2} = I_{total} \times \omega\) where \(I_{total}\) is the total moment of inertia of the bullet-rod system and \(\omega\) is the angular velocity after the collision. To find the \(I_{total}\), we need to find the moment of inertia of the rod (\(I_R\)) and the bullet (\(I_B\)) and add them. The moment of inertia of the rod can be found by the formula of a rod pivoted about its center: \(I_R = \frac{1}{12} m_{R} L^2\) The moment of inertia of the bullet can be found by the following formula: \(I_B = m_{B} r^2\) Now, we can find the total moment of inertia: \(I_{total} = I_{R} + I_{B} = \frac{1}{12} m_{R} L^2 + m_{B} r^2\) Plugging in the values, we get: \(I_{total} = \frac{1}{12} (5.00 \mathrm{kg})(1.00 \mathrm{m})^2 + (1.00 \times 10^{-2} \mathrm{kg}) (\frac{1}{4} \mathrm{m})^2 = 0.4167 \mathrm{kg \cdot m^2}\)

Find the angular velocity after the collision

Now that we have the total moment of inertia, we can find the angular velocity: \(\omega = \frac{L_{2}}{I_{total}}\) Using the values we have calculated earlier: \(\omega = \frac{0.25 \mathrm{kg \cdot m^2/s}}{0.4167 \mathrm{kg \cdot m^2}} \approx 0.60 \mathrm{rad/s}\) This is the angular velocity of the bullet-rod system after the collision.

Calculate the initial kinetic energy

The initial kinetic energy of the bullet can be found using the formula: \(KE_{1} = \frac{1}{2} m_{B} v^2\) Plugging in the values: \(KE_{1} = \frac{1}{2} (1.00 \times 10^{-2} \mathrm{kg}) (100 \mathrm{m/s})^2 = 50.00 \mathrm{J}\)

Calculate the final kinetic energy

The final kinetic energy of the bullet-rod system can be found using the formula: \(KE_{2} = \frac{1}{2} I_{total} \omega^2\) Plugging in the values: \(KE_{2} = \frac{1}{2} (0.4167 \mathrm{kg \cdot m^2}) (0.60 \mathrm{rad/s})^2 \approx 7.50 \mathrm{J}\)

Find the loss in kinetic energy

Finally, we can find the loss in kinetic energy during the collision: \(\Delta KE = KE_{1} - KE_{2}\) \(\Delta KE = 50.00 \mathrm{J} - 7.50 \mathrm{J} = 42.50 \mathrm{J}\) The loss in kinetic energy is 42.50 Joules during the collision. In this exercise, we have found the angular velocity \(\omega\) of the bullet-rod system after the collision to be approximately \(0.60 \, \mathrm{rad/s}\) and the loss in kinetic energy during the collision to be \(42.50 \, \mathrm{J}\).

Key concepts

Moment of Inertia

The concept of moment of inertia is central to understanding rotational motion. It is the rotational equivalent of mass for linear motion, suggesting how resistant an object is to change in its rotational state. The moment of inertia depends on the distribution of mass around the axis of rotation. The further the mass is from the axis, the higher the moment of inertia, and the more torque is required to change its rotational speed.

For different shapes and mass distributions, the moment of inertia has different formulas. In the exercise, we examined a composite system consisting of a rod and a bullet embedded into it. The rod's moment of inertia is given by \(I_R = \frac{1}{12} m_R L^2\), which represents a rod pivoted about its center. The bullet, treated as a point mass at a distance \(r\), contributes \(I_B = m_B r^2\) to the system's total moment of inertia. Add these two together, and we get the total moment of inertia, which is crucial for analyzing the rotational motion post-collision.

Angular Velocity

Angular velocity, often denoted by \(\backslash omega\), is a measure of how fast an object rotates or revolves relative to another point, usually the center of a rotational system. It tells us the rate at which the angle is changing over time and is measured in radians per second (\(\backslash \text {rad/s}\)).

In our exercise, once the bullet embeds itself into the rod, the entire bullet-rod system rotates with a new constant angular velocity. By applying the conservation of angular momentum, we were able to calculate the angular velocity of this system after the collision. Knowing the total moment of inertia and the initial angular momentum before the collision, the angular velocity is found by dividing the angular momentum by the moment of inertia: \(\backslash omega = \frac{L_{2}}{I_{total}}\).

Kinetic Energy Loss

When we talk about kinetic energy loss in a system, we're referring to the difference between the initial kinetic energy and the final kinetic energy after an event, such as a collision. Kinetic energy is the energy of motion, and it's calculated for a rotating object using \(\frac{1}{2} I \backslash omega^2\), where \(I\) is the moment of inertia and \(\backslash omega\) is the angular velocity.

In an inelastic collision, like the one in our exercise where the bullet embeds in the rod, some kinetic energy is converted into other forms of energy, such as heat or sound, and is therefore 'lost' from the system. By comparing the initial kinetic energy of the bullet and the final kinetic energy of the bullet-rod system, we can determine the amount of kinetic energy lost during the collision.

Rotational Dynamics

Rotational dynamics is the study of the forces and torques that cause changes in rotational motion, governed by principles similar to Newton's laws of motion for translational dynamics. In this realm, torque is analogous to force, and moment of inertia is analogous to mass. The conservation of angular momentum, a fundamental principle of rotational dynamics, states that if no external torque acts on a system, the total angular momentum remains constant over time.

Through the lens of rotational dynamics, we analyzed a collision in which a bullet, initially moving linearly, became part of a rotating system involving a rod. This exercise required us to apply conservation laws and concepts of rotational dynamics, such as torque, moment of inertia, and angular velocity, to solve for the post-collision rotational state and energy loss of the system.