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Chapter 10: Problem 80

A \(5.00 \mathrm{~kg}\) ball is dropped from a height of \(12.0 \mathrm{~m}\) above one end of a uniform bar that pivots at its center. The bar has mass 8.00 \(\mathrm{kg}\) and is \(4.00 \mathrm{~m}\) in length. At the other end of the bar sits another 5.00 \(\mathrm{kg}\) ball, unattached to the bar. The dropped ball sticks to the bar after the collision. How high will the other ball go after the collision?

Short Answer

Expert verified
The other ball will go up to a height of 3.33 meters after the collision.

Step by step solution

01

Understand the problem

In this problem a ball is dropped from a height onto a bar, which then forces another ball to move upward. The key information given includes the weights of the balls, the length of the bar, and its weight. The question asks for how high the other ball will go after the collision. The problem involves energy conservation principles, which involve kinetic and potential energy.
02

Visualizing the problem

Visualize this setup as a see-saw system. When the first ball falls, it hits one end of the bar and sticks to it. This raises the other ball to a certain height, which we need to find.
03

Applying the principle of energy conservation

According to the law of the conservation of energy, the total energy of an isolated system remains constant over time. The initial kinetic energy of the ball before the collision will add up to the potential energy at the maximum height of the ball after the collision. Initially, the falling ball has potential energy which turns into kinetic energy right before the collision. We can represent this as \(PE_{initial} = KE_{final}\), where PE is potential energy and KE is kinetic energy.
04

Calculating the initial Potential Energy

The potential energy can be calculated as \(PE = mgh\), where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height. The mass of the ball is 5 kg and it is dropped from a height of 12 m, so the initial potential energy is \(PE_{initial}=5*9.8*12 = 588 J\), which becomes completely kinetic at the bottom just before the collision.
05

Calculating the final kinetic energy and the maximum height

According to the conservation of energy, the kinetic energy just before the collision will convert into total potential energy at the maximum height after collision. This potential energy can be calculated as \(PE_{final}= m'*g*h'\), where m' is the new mass (8 kg bar + 5 kg dropped ball + 5 kg the other ball = 18 kg) and we solve for \(h'\) (the height we are looking for), \(h' = PE_{final}/(m'*g) = 588/(18*9.8)\)
06

Providing the final answer

Calculate \(h' = 588/(18*9.8) = 3.33\) meters. This is the height to which the other ball will rise after the first one hits the bar.

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Key Concepts

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

Kinetic Energy
Kinetic Energy is the energy possessed by an object due to its motion. Imagine pushing a ball and watching it glide across the floor. That's kinetic energy in action! It is calculated using the equation \[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass of the object and \( v \) is its velocity.

In the exercise, the ball initially has potential energy due to its height. As it falls, this potential energy is transformed into kinetic energy just before the collision with the bar. Such transformation of energy is crucial in analyzing motion.
  • Increases with mass: A heavier object will have more kinetic energy at the same speed compared to a lighter one.
  • Increases with speed: Even a small object can have significant kinetic energy if it's moving fast enough.
Understanding kinetic energy helps you figure out what happens during the collision 鈥� how energy gets transferred to the bar and moves the other ball.
Potential Energy
Potential Energy is the stored energy of an object due to its position or state. A classic example is a ball held at a height, ready to fall. The potential energy is given by the formula:
\[ PE = mgh \]where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (9.8 m/s虏 on Earth), and \( h \) is the height above the ground.

In the exercise scenario, the potential energy of the ball at the start is 588 J, resulting from its mass (5 kg) and height (12 m). This energy is released when the ball falls, allowing us to calculate its subsequent motion through the system.
  • Affected by height: The higher the object, the greater its potential energy.
  • Dependant on gravity: More gravity means more potential energy.
Potential energy provides insight into the capability of an object to do work, such as moving another object like the bar in the problem.
Rotational Dynamics
Rotational Dynamics studies the motion of objects that rotate. Picture a turning bike wheel or spinning top. This area of physics looks at how and why things spin, influenced by forces and torques. In our exercise, the pivoting of the bar showcases rotational motion.

When the ball hits one end of the bar, it creates a torque that causes the bar to rotate. This rotational action depends on three factors:
  • Moment of inertia: The resistance of the bar to change its rotational motion, influenced by its mass and length.
  • Torque: The force applied at a distance from the pivot, causing rotation.
  • Angular momentum: Conserved when no external torques act, linking linear velocity of the ball to rotational velocity.
Rotational dynamics highlight how energy from a falling object is transferred to enable the bar's motion, causing the other ball to rise. By analyzing these aspects, you grasp how rotational motion and energy conservation work together.

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Most popular questions from this chapter

A solid wood door \(1.00 \mathrm{~m}\) wide and \(2.00 \mathrm{~m}\) high is hinged along one side and has a total mass of \(40.0 \mathrm{~kg}\). Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass \(0.500 \mathrm{~kg}\), traveling perpendicular to the door at \(12.0 \mathrm{~m} / \mathrm{s}\) just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

A cord is wrapped around the rim of a solid uniform wheel \(0.250 \mathrm{~m}\) in radius and of mass \(9.20 \mathrm{~kg} .\) A steady horizontal pull of \(40.0 \mathrm{~N}\) 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?

A hollow, thin-walled sphere of mass \(12.0 \mathrm{~kg}\) and diameter \(48.0 \mathrm{~cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value \(1.10 .\) (a) What are the units of the constants \(A\) and \(B ?\) (b) At the time \(3.00 \mathrm{~s}\), find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

An engine delivers 175 hp to an aircraft propeller at 2400 rev \(/\) min. (a) How much torque does the aircraft engine provide? (b) How much work does the engine do in one revolution of the propeller?
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