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Magazine Chapter 11: Problem 9
A solid ball rolls smoothly from rest (starting at height \(H=6.0 \mathrm{~m}\) ) until it leaves the horizontal section at the end of the track, at height \(h=2.0 \mathrm{~m}\). How far horizontally from point \(A\) does the ball hit the floor?
Short Answer
Expert verified
The ball hits the floor 4.79 meters horizontally from point A.
Step by step solution
01
Conservation of Energy
Since the solid ball rolls smoothly and conserves mechanical energy, we can use the conservation of energy to determine its velocity at height \( h = 2.0 \text{ m} \). Initially, the ball has potential energy (PE) and no kinetic energy (KE) since it starts from rest. When it reaches height \( h \), some of this energy is converted into kinetic energy.The potential energy at height \( H \) is \( PE = m g H \), and at height \( h \) is \( PE = m g h \).By conservation of energy:\[ m g H = m g h + \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \]Where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a solid ball, \( I = \frac{2}{5} m r^2 \) and \( \omega = \frac{v}{r} \).Substitute \( I \) and \( \omega \):\[ m g H = m g h + \frac{1}{2} m v^2 + \frac{1}{2} \left( \frac{2}{5} m r^2 \right) \left( \frac{v}{r} \right)^2 \]Simplifying gives:\[ m g H = m g h + \frac{7}{10} m v^2 \]Canceling \( m \) from both sides, we have:\[ g H = g h + \frac{7}{10} v^2 \]Thus:\[ v^2 = \frac{10}{7} g (H - h) \]
02
Calculate the Velocity Before Leaving the Track
Substitute the known values of \( g = 9.8 \text{ m/s}^2 \), \( H = 6.0 \text{ m} \), and \( h = 2.0 \text{ m} \) into the equation for \( v^2 \):\[ v^2 = \frac{10}{7} \times 9.8 \times (6.0 - 2.0) \]\[ v^2 = \frac{10}{7} \times 9.8 \times 4 \]\[ v^2 = \frac{10}{7} \times 39.2 \]\[ v^2 = 56 \]Therefore, the velocity \( v \) is:\[ v = \sqrt{56} \approx 7.48 \text{ m/s} \]
03
Analyze the Projectile Motion
Once the ball leaves the track, it acts as a projectile. We need to determine how far horizontally it travels before hitting the ground.We use projectile motion equations for horizontal distance \( x \) by calculating the time \( t \) it takes to fall the remaining \( 2.0 \text{ m} \).Using: \( h = \frac{1}{2} g t^2 \)Set \( h = 2.0 \text{ m} \), so:\[ 2 = \frac{1}{2} \times 9.8 \times t^2 \]\[ 2 = 4.9 t^2 \]Solve for \( t \):\[ t^2 = \frac{2}{4.9} \]\[ t^2 \approx 0.41 \]\[ t \approx 0.64 \text{ seconds} \]
04
Horizontal Distance Calculation
With the time \( t \) and the horizontal velocity \( v \), determine the distance \( x \).\[ x = v \times t \]\[ x = 7.48 \times 0.64 \]\[ x \approx 4.79 \text{ m} \]Thus, the horizontal distance from point \( A \) to where the ball hits the ground is \( 4.79 \) meters.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rolling Motion
Rolling motion occurs when an object rotates around its own axis while it moves linearly along a surface. This blend of translational and rotational movement is seamless and is seen in a variety of everyday phenomena, like a ball rolling down a slope. To understand rolling motion, consider a solid ball that begins its journey at rest from a certain height. As the ball rolls down, it converts its potential energy into kinetic energy. This kinetic energy is divided into translational kinetic energy, due to its motion along the track, and rotational kinetic energy, due to its spinning. A key factor is the ball's moment of inertia, which defines how mass is distributed and affects the ball's rolling speed. In rolling without slipping, the point of contact between the ball and the ground is momentarily at rest with respect to the ground. This "rolling friction" allows the ball to maintain energy efficiency, governing how quickly it can move.
Projectile Motion
Once our ball, now carrying the kinetic energy from rolling, leaves the edge of the track, it transitions into projectile motion. This type of motion involves the ball moving through the air under the influence of gravity alone. Understanding projectile motion involves breaking the ball's path into horizontal and vertical components. While gravity pulls the ball downward, the horizontal motion remains unaffected by gravitational forces. The time it takes for the ball to hit the ground is identical to the time it would take if it just fell vertically from the same height. Using this time, and knowing the horizontal speed (obtained from earlier kinetic energy calculations), we can determine how far the ball travels horizontally. Key formulas for projectile motion often include:
- Height: Using \[ h = \frac{1}{2} g t^2 \], the fall time is determined for a given height.
- Horizontal Distance: Calculated as the initial velocity times the fall time \[ x = v_x imes t \].
Mechanical Energy
Mechanical energy is a broad term that combines both kinetic and potential energy, making it key in understanding energy conservation in physics. For an object like a solid ball rolling down a hill, mechanical energy is constantly transformed between potential and kinetic forms without any loss, when we ignore factors like air resistance and friction. Initially, the ball has potential energy, which depends on its mass, gravitational acceleration, and height (\( PE = mgh \)). As it rolls down, this energy becomes kinetic, reflecting in the speed of the ball. The kinetic energy (\( KE = \frac{1}{2}mv^2 \)) for rolling objects has two components:
- Translational kinetic energy, related to the linear motion.
- Rotational kinetic energy, related to the spin of the ball.
Moment of Inertia
The moment of inertia is a fundamental concept that helps us quantify an object's resistance to changes in its rotational motion. Think of it as the equivalent of mass in rotational equations. For a solid ball, the moment of inertia is determined by the distribution of its mass around the axis of rotation, calculated as\( I = \frac{2}{5}mr^2 \). This value influences how the ball accelerates as it rolls down. A larger moment of inertia means more difficulty in starting to spin. In our example, the formula for potential and kinetic energy consideration uses the moment of inertia to account for the energy linked with the ball's spin \( \frac{1}{2}I\omega^2 \). Because \( \omega = \frac{v}{r} \), the rotational speed simplifies to depend on the linear speed of the object. Understanding moment of inertia is vital in dynamic systems, where calculating the correct energy transformation is crucial.
