Problem 47 - University Physics with Modern Physics

Question

A small 10.0-g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50.0 g and is 100 cm in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 cm/s relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps? (b) What is the total kinetic energy of the system just after the bug leaps? (c) Where does this energy come from?

Step-by-Step Solution

Answer

(a) \( \omega \approx 0.12 \, \text{rad/s} \), (b) total KE \( \approx 3.2 \times 10^{-4} \, \text{J} \), (c) energy from bug's muscle work.

Step 1: Understanding the Conservation of Angular Momentum

Since the system is initially at rest and there is no external torque, the initial angular momentum of the system is zero. This implies that the total angular momentum of the bar and the bug after the jump should still be zero. Given the mass of the bug (10 g) and its velocity (20 cm/s), calculate the bug's angular momentum relative to the pivot point as: \( L_{bug} = m_{bug} \cdot v_{bug} \cdot r \), where \( r \) is the distance from the pivot.

Step 2: Calculating Angular Momentum of the Bug

Calculate the bug's angular momentum using its mass (converted to kg, 0.01 kg), its speed (converted to m/s, 0.2 m/s), and the length of the bar (converted to meters, 1.0 m). \( L_{bug} = 0.01 \cdot 0.2 \cdot 1.0 = 0.002 \, \text{kg} \cdot \text{m}^2/\text{s} \).

Step 3: Applying Angular Momentum Conservation

Since the initial angular momentum is zero and the only moving part initially was the bug, after the bug jumps, the bar should have an equal and opposite angular momentum to conserve angular momentum: \( L_{bar} = -L_{bug} = -0.002 \, \text{kg} \cdot \text{m}^2/\text{s} \).

Step 4: Calculate Angular Speed of the Bar

The moment of inertia \( I \) of a uniform bar pivoting at one end is \( \frac{1}{3} M L^2 \), where \( M = 0.05 \, \text{kg} \) and \( L = 1 \, \text{m} \). Calculate \( I: I = \frac{1}{3} \times 0.05 \times 1^2 = 0.0167 \, \text{kg} \cdot \text{m}^2 \). The angular speed \( \omega \) is given by \( \omega = \frac{L_{bar}}{I} = \frac{-0.002}{0.0167} \approx -0.12 \, \text{rad/s} \).

Step 5: Calculating the Total Kinetic Energy After the Jump

Calculate the kinetic energy of the bar using \( KE_{bar} = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.0167 \times (0.12)^2 \) which gives \( \approx 1.2 \times 10^{-4} \, \text{J} \). Calculate the kinetic energy of the bug using \( KE_{bug} = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.01 \times (0.2)^2 \approx 2.0 \times 10^{-4} \, \text{J} \). The total kinetic energy is the sum of the bug's and bar's: \( \approx 3.2 \times 10^{-4} \, \text{J} \).

Step 6: Identifying the Source of this Energy

The energy comes from the work done by the bug as it jumps off the bar, converting its internal energy (muscle power) into kinetic energy of the bug and the bar.

Key Concepts

Angular Speed

Angular speed is a measure of how quickly an object rotates around an axis. It is represented by the symbol \( \omega \) and is typically measured in radians per second (rad/s). In this exercise, the angular speed is determined by how the bar rotates after the bug jumps off.
When dealing with problems involving angular momentum, like this one, it's essential to use the principle of conservation of angular momentum.
This principle states that if no external torque acts on a system, the total angular momentum of the system remains constant.

• When the system is initially at rest, and the bug jumps, the angular momentum before and after must be the same.
• The bug's movement creates angular momentum, and the bar responds by rotating with a specific angular speed that balances the system.

The bar's angular speed \( \omega \) is calculated using the formula \( \omega = \frac{L_{bar}}{I} \).
Here, \( L_{bar} \) is the angular momentum of the bar and \( I \) is the moment of inertia.
This formula helps us understand how changes in angular momentum affect rotational speed.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It is a crucial concept when studying systems like the bug and the bar.

• Kinetic energy of an object is given by \( KE = \frac{1}{2} m v^2 \) for linear motion or \( KE = \frac{1}{2} I \omega^2 \) for rotational motion.
• In this scenario, both the bug and the bar have kinetic energies after the jump, which need to be summed up to find the total kinetic energy of the system.

In this exercise, the bug gains kinetic energy by using its muscles to leap off the bar.
The bar, in response, gains rotational kinetic energy due to its angular speed.
The combination of these energies gives us the total kinetic energy of the system.
The calculation of this total kinetic energy involves determining both the translational kinetic energy of the jumping bug and the rotational kinetic energy of the bar.

Moment of Inertia

The moment of inertia is a fundamental concept in rotational dynamics.
Think of it as the rotational analogue of mass for linear motion. It quantifies an object's resistance to changes in its rotational motion.

• The moment of inertia \( I \) of an object depends on its mass and the distribution of that mass relative to the axis of rotation.
• For a thin bar rotating about an end, the moment of inertia is calculated using \( I = \frac{1}{3} M L^2 \).

In this problem, the bar's moment of inertia affects how it rotates after the bug jumps.
The bar's mass and length both play a role in determining its moment of inertia.
Understanding moment of inertia is key to predicting the motion of rotating objects.
It allows one to calculate the angular speed and kinetic energies that result from changes in motion, like those seen after the bug's jump.