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Chapter 10: Problem 49

A thin, uniform metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10.0 m/s. The ball rebounds in the opposite direction with a speed of 6.00 m/s. (a) Find the angular speed of the bar just after the collision. (b) During the collision, why is the angular momentum conserved but not the linear momentum?

Short Answer

Expert verified
The angular speed of the bar is approximately 5.89 rad/s. Angular momentum is conserved due to the absence of external torques.

Step by step solution

01

Identify the Initial and Final States

This is an inelastic collision problem involving a ball and a bar. Before the collision, only the ball has linear momentum, moving towards the bar. After the collision, the ball has rebounded, and the bar gains angular velocity.
02

Calculate Initial Angular Momentum

The initial linear momentum of the ball is given by \( p_{i} = m \cdot v_{i} \), where \( m = 3.00 \text{ kg} \) and \( v_{i} = 10.0 \text{ m/s} \). Hence, \( p_{i} = 3.00 \cdot 10.0 = 30.0 \text{ kg m/s} \). The initial angular momentum \( L_{i} \) about the pivot is \( L_{i} = r \cdot p_{i} = 1.50 \cdot 30.0 = 45.0 \text{ kg m}^2/\text{s} \).
03

Calculate Final Angular Momentum

After the collision, the final linear momentum of the ball is \( p_{f} = m \cdot (-v_{f}) \), where \( v_{f} = 6.00 \text{ m/s} \) in the opposite direction. Thus, \( p_{f} = 3.00 \cdot (-6.00) = -18.0 \text{ kg m/s} \). The final angular momentum of the ball about the pivot is \( L_{f, ball} = r \cdot p_{f} = 1.50 \cdot (-18.0) = -27.0 \text{ kg m}^2/\text{s} \).
04

Apply the Conservation of Angular Momentum

The total initial angular momentum \( L_{i} \) must equal the total final angular momentum \( L_{f} = L_{f, ball} + L_{f, bar} \). Hence, \( 45.0 = -27.0 + L_{f, bar} \). Thus, \( L_{f, bar} = 72.0 \text{ kg m}^2/\text{s} \).
05

Calculate Angular Speed of the Bar

The moment of inertia \( I \) of the bar about the pivot is given by \( I = \frac{1}{3} mL^2 \). Here, \( m = \frac{90.0}{9.81} \approx 9.17 \text{ kg} \) and \( L = 2.00 \text{ m} \). Thus, \( I = \frac{1}{3} \times 9.17 \times 2^2 = 12.22 \text{ kg m}^2 \).\( L_{f, bar} = I \cdot \omega \), therefore, \( \omega = \frac{L_{f, bar}}{I} = \frac{72.0}{12.22} \approx 5.89 \text{ rad/s} \).
06

Reason for Conservation of Angular Momentum

Angular momentum is conserved because no external torques act on the system. Linear momentum is not conserved because there is an external force from the pivot, which applies a torque affecting linear momentum but not angular momentum.

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Key Concepts

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

Inelastic Collision
An inelastic collision is a type of collision in which the colliding objects stick together or move with some different velocities after impact. Although the total kinetic energy is not conserved due to the dissipation of energy in the form of heat, sound, or deformation, the total momentum is conserved.
In our exercise, a 3.00-kg ball hits a hanging metal bar. During the collision, the ball changes direction, indicating an inelastic property since kinetic energy is not conserved.
This conclusively demonstrates a real-world example of an inelastic collision, where motion and some energy transform within the structures involved.
Moment of Inertia
The moment of inertia is an essential concept in rotational mechanics. It measures an object's resistance to changes in its rotation rate. You can think of it as the rotational equivalent of mass in linear motion.
In the exercise, the moment of inertia of the bar is calculated to see how it impacts the bar's angular velocity post-collision. Using the formula: \[ I = \frac{1}{3} m L^2 \] where \( m \) is the mass of the bar and \( L \) is its length, we find the bar's moment of inertia is 12.22 kg m虏.
This is crucial because it helps determine how the impact from the ball affects the bar鈥檚 rotation speed.
Conservation of Momentum
In every collision, like the one between the ball and the bar, the principle of conserving momentum plays a pivotal role. This principle states that the total momentum before the collision must equal the total momentum after, as long as no external forces act upon the system.
Even though the linear momentum of our system isn't conserved due to the external pivot force, the angular momentum is conserved. This is because no external torque is acting on the system during the collision.
  • Total initial angular momentum: 45.0 kg m虏/s
  • Final angular momentum shared by the rebounding ball and spinning bar: 45.0 kg m虏/s
This conservation ensures we can accurately solve for post-collision angular velocity.
Angular Velocity
Angular velocity refers to how fast an object spins around an axis. This is a direct outcome of the angular momentum for a rotating system like our metal bar.
In our case, post-collision, the bar has gained angular velocity, which we calculate using:\[ \omega = \frac{L_{f, \text{bar}}}{I} \]where \( L_{f, \text{bar}} \) is the final angular momentum of the bar and \( I \) is the moment of inertia. This results in an angular velocity of approximately 5.89 rad/s.
Understanding this calculation shows how the change in form of momentum manifests as a pace in the object's rotation after being struck by another moving object.

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Most popular questions from this chapter

You complain about fire safety to the landlord of your high-rise apartment building. He is willing to install an evacuation device if it is cheap and reliable, and he asks you to design it. Your proposal is to mount a large wheel (radius 0.400 m) on an axle at its center and wrap a long, light rope around the wheel, with the free end of the rope hanging just past the edge of the roof. Residents would evacuate to the roof and, one at a time, grasp the free end of the rope, step off the roof, and be lowered to the ground below. (Ignore friction at the axle.) You want a 90.0-kg person to descend with an acceleration of g/4. (a) If the wheel can be treated as a uniform disk, what mass must it have? (b) As the person descends, what is the tension in the rope?

A 392-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is 0.600 m, and its moment of inertia about its rotation axis is 0.800MR\(^2\). Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 2600 J. Calculate \(h\).

A hollow, spherical shell with mass 2.00 kg rolls without slipping down a 38.0\(^\circ\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to 4.00 kg?

A grindstone in the shape of a solid disk with diameter 0.520 m and a mass of 50.0 kg is rotating at 850 rev/min. You press an ax against the rim with a normal force of 160 N (Fig. P10.54), and the grindstone comes to rest in 7.50 s. Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axle exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?
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