Problem 65 - Fundamentals of Physics

Question

Two \(2.00 \mathrm{~kg}\) balls are attached to the ends of a thin rod of length \(50.0 \mathrm{~cm}\) and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (Fig.\(11-57),\) a \(50.0 \mathrm{~g}\) wad of wet putty drops onto one of the balls, hitting it with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) and then sticking to it. (a) What is the angular speed of the system just after the putty wad hits? (b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before? (c) Through what angle will the system rotate before it momentarily stops?

Step-by-Step Solution

Answer

(a) 0.148 rad/s, (b) 0.0123, (c) 0.0118 radians.

Step 1: Calculate Initial Linear Momentum

The momentum of the putty wad before collision is given by its mass times its velocity. Let \( m_1 = 0.05 \, \text{kg} \) (mass of putty) and \( v = 3.00 \, \text{m/s} \) (velocity of putty). Therefore, the initial linear momentum \( p_i = m_1 \cdot v = 0.05 \, \text{kg} \times 3.00 \, \text{m/s} = 0.15 \, \text{kg} \cdot \text{m/s} \).

Step 2: Convert Linear Momentum to Angular Momentum

As the putty sticks to the ball, its linear momentum becomes angular momentum about the axis through the center. The distance to the axis is half the length of the rod, \( r = 0.25 \, \text{m} \). Thus, the angular momentum is \( L_i = r \cdot p_i = 0.25 \, \text{m} \times 0.15 \, \text{kg} \cdot \text{m/s} = 0.0375 \, \text{kg} \cdot \text{m}^2/\text{s} \).

Step 3: Calculate Moment of Inertia after Collision

The moment of inertia \( I \) of the system after the collision includes the two balls and the putty. For each ball, \( I_b = m_2 \cdot r^2 = 2.00 \, \text{kg} \times (0.25 \, \text{m})^2 = 0.125 \, \text{kg} \cdot \text{m}^2 \). The moment of inertia for the putty is similar: \( I_p = m_1 \cdot r^2 = 0.05 \, \text{kg} \times (0.25 \, \text{m})^2 = 0.003125 \, \text{kg} \cdot \text{m}^2 \). The total \( I = 2 \cdot 0.125 + 0.003125 = 0.253125 \, \text{kg} \cdot \text{m}^2 \).

Step 4: Calculate Angular Speed after Collision

Using conservation of angular momentum, the angular speed \( \omega \) just after the collision is \( \omega = \frac{L_i}{I} = \frac{0.0375}{0.253125} = 0.148 \, \text{rad/s} \).

Step 5: Calculate Initial Kinetic Energy of the Putty

The initial kinetic energy of the putty is given by \( KE_i = \frac{1}{2} m_1 v^2 = \frac{1}{2} \times 0.05 \times 3^2 = 0.225 \, \text{J} \).

Step 6: Calculate Final Kinetic Energy of the System

The kinetic energy of the system after the collision is \( KE_f = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.253125 \times (0.148)^2 = 0.00277 \, \text{J} \).

Step 7: Ratio of Kinetic Energy after to before Collision

The ratio is \( \frac{KE_f}{KE_i} = \frac{0.00277}{0.225} \approx 0.0123 \).

Step 8: Calculate Angle of Rotation to Stop

The system will convert kinetic energy to potential energy as it rises. Using conservation of energy, \( \frac{1}{2} I \omega^2 = mgh \). \( h \) is the vertical drop related to angle \( \theta \), \( h = r(1-\cos \theta) \). Solving for \( \theta \), based on energy conversion, leads to \( \theta = \cos^{-1}(1-\frac{\omega^2 r}{2g}) \). Substitute known values to find \( \theta \approx 0.0118 \, \text{radians}\).

Key Concepts

Conservation of Angular Momentum

Angular momentum plays a crucial role in rotational dynamics, acting as the rotational equivalent of linear momentum. It is a measure of the angular motion of a system and is defined as the product of an object's moment of inertia and its angular velocity. The principle of conservation of angular momentum states that if no external torque acts on a system, its total angular momentum remains constant. This fundamental idea is applicable in various scenarios, from spinning planets to rotating ice skaters.

In our exercise involving the putty and rod system, the law of conservation of angular momentum is key to determining the angular speed after the putty sticks to the ball. Since the rod is rotating in a frictionless environment, the total angular momentum before the collision equals the total angular momentum after the collision.

• Initial angular momentum: dependent on the putty's linear momentum turning into angular momentum when it becomes part of the rotating system.
• Final angular momentum: calculated taking into account the combined moment of inertia of the balls and putty.

Understanding this principle allows us to seamlessly transition from before the collision - where the putty has linear momentum - to the aftermath, where the entire system is in rotational motion.

Moment of Inertia

The moment of inertia is an object's resistance to changes in its rotational motion and is dependent on the mass distribution relative to the axis of rotation. It is pivotal in rotational dynamics, as it influences how easily an object can begin rotating or change its rotational speed. The moment of inertia for point masses can be computed using the formula:\[ I = \sum m_ir_i^2 \]where \(m_i\) is the mass and \(r_i\) is the distance from the axis.

In the exercise, we derive the moment of inertia for the rod system by individually calculating for the two identical balls and the putty. The rod itself, described as having negligible mass, doesn’t contribute to the moment of inertia.

• For each ball: it depends on the mass of the ball and its distance from the axis, leading to \( I_b = m_2 \cdot r^2 \).
• For the putty: similarly, as it adheres to a ball at a distance \(r\), contributing its own \( I_p = m_1 \cdot r^2 \).
• Total moment of inertia: summing these values gives the combined resistance to angular acceleration for both balls and the putty.

This property not only determines rotational acceleration but also affects how kinetic energy is distributed across the system post-collision.

Kinetic Energy

Kinetic energy in rotational contexts is analogous to that in linear motion but involves rotational motion parameters. The rotational kinetic energy of an object is given by the equation:\[ KE = \frac{1}{2} I \omega^2 \]where \(I\) is the moment of inertia and \(\omega\) is the angular speed. After a collision, we can observe how kinetic energy transforms, which is vital in understanding energy efficiency and loss during interactions.

In this example, the initial kinetic energy belonged solely to the putty but transformed post-collision as it adheres to the ball and rotates within the system. Calculations show the kinetic energy ratio before and after collision, illustrating how much energy is retained within the system post-impact, shedding light on conversion efficiency:

• Initial kinetic energy: from the putty's velocity before impact, calculated using \(\frac{1}{2} m_1 v^2\).
• Final kinetic energy: based on the combined moment of inertia and the newly calculated angular speed, demonstrating transformation efficiency.
• Energy ratio: indicates the conservation of energy extent, providing insights into energy losses during momentum conversion.

Understanding these transformations helps reveal the nuances of rotational motion and energy management in rotating systems.