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Chapter 10: Problem 47

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?

Short Answer

Expert verified
(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 by step solution

01

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.
02

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). \( L_{bug} = 0.01 \cdot 0.2 \cdot 1.0 = 0.002 \, \text{kg} \cdot \text{m}^2/\text{s} \).
03

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: \( L_{bar} = -L_{bug} = -0.002 \, \text{kg} \cdot \text{m}^2/\text{s} \).
04

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 \( \omega = \frac{L_{bar}}{I} = \frac{-0.002}{0.0167} \approx -0.12 \, \text{rad/s} \).
05

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: \( \approx 3.2 \times 10^{-4} \, \text{J} \).
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.650 m in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects 0.200 m at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is 8.00 kg; its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at 5.00 rev/s. Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at 0.050 rev/s; (c) when the shaft is rotating in a horizontal plane about its center at 0.300 rev/s. (d) At what rate must the shaft rotate in order that it may be supported at one end only?

A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00-m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The 0.800-kg rod has a light basket at the other end to catch the 0.163-kg puck. The puck slides across the ice with velocity \(\overrightarrow{v}\) (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates. If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod?

A solid ball is released from rest and slides down a hillside that slopes downward at 65.0\(^\circ\) from the horizontal. (a) What minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping to occur? (b) Would the coefficient of friction calculated in part (a) be sufficient to prevent a hollow ball (such as a soccer ball) from slipping? Justify your answer. (c) In part (a), why did we use the coefficient of static friction and not the coefficient of kinetic friction?

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_x\) and \(a_z\) are approximately zero and \(v_x\) and \(\omega_z\) are approximately constant. Rolling without slipping means \(v_x = r\omega_z\) and \(a_x = r\alpha_z\) . If an object is set in motion on a surface \(without\) these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\), rotating with angular speed \(\omega_0\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_k\). (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_x\) of the center of mass and \(a_z\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_z = \omega_0\) but \(v_x =\) 0. Rolling without slipping sets in when \(v_x = r\omega_z\) . Calculate the \(distance\) the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

We can roughly model a gymnastic tumbler as a uniform solid cylinder of mass 75 kg and diameter 1.0 m. If this tumbler rolls forward at 0.50 rev/s, (a) how much total kinetic energy does he have, and (b) what percent of his total kinetic energy is rotational?
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