Problem 7 - Fundamentals of Physics

Question

A solid cylinder of radius \(10 \mathrm{~cm}\) and mass \(12 \mathrm{~kg}\) starts from rest and rolls without slipping a distance \(L=6.0 \mathrm{~m}\) down a roof that is inclined at angle \(\theta=30^{\circ}\) (a) What is the angular speed of the cylinder about its center as it leaves the roof? (b) The roof's edge is at height \(H=5.0 \mathrm{~m}\). How far horizontally from the roof's edge does the cylinder hit the level ground?

Step-by-Step Solution

Answer

(a) The angular speed is approximately 62.79 rad/s. (b) The horizontal distance is about 3.48 m.

Step 1: Calculate the Potential Energy

The cylinder starts from rest, implying its initial kinetic energy is zero. Use the potential energy formula, \[ PE = mgh \]where \( m = 12 \mathrm{~kg} \), \( g = 9.81 \mathrm{~m/s^2} \), and \( h = L\sin\theta \) with \( L = 6.0 \mathrm{~m} \) and \( \theta = 30^{\circ} \). Calculate \( h \) by \[ h = 6.0 \mathrm{~m} \times \sin(30^{\circ}) = 6.0 \mathrm{~m} \times 0.5 = 3.0 \mathrm{~m} \]Now, calculate the potential energy:\[ PE = 12 \mathrm{~kg} \times 9.81 \mathrm{~m/s^2} \times 3.0 \mathrm{~m} = 353.16 \mathrm{~J} \]

Step 2: Apply Energy Conservation

The potential energy is converted into translational and rotational kinetic energy. The conservation of energy equation is:\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]The moment of inertia \( I \) for a solid cylinder is \( \frac{1}{2}mr^2 \). Use this and solve for \( \omega \), knowing \( v = r\omega \): \[ 353.16 \mathrm{~J} = \frac{1}{2} \times 12 \mathrm{~kg} \times v^2 + \frac{1}{2} \times \frac{1}{2} \times 12 \mathrm{~kg} \times (0.1 \mathrm{~m})^2 \times \omega^2 \]This simplifies to \[ 353.16 \mathrm{~J} = 6v^2 + 0.03\omega^2 \]Substituting \( v = 0.1\omega \) gives:\[ 353.16 = 6(0.1\omega)^2 + 0.03\omega^2 \]\[ 353.16 = 0.06\omega^2 + 0.03\omega^2 \]\[ 353.16 = 0.09\omega^2 \]Solve for \( \omega \):\[ \omega = \sqrt{\frac{353.16}{0.09}} \approx 62.79 \mathrm{~rad/s} \]

Step 3: Calculate the Time of Flight

The vertical drop is the height difference \( H = 5.0 \mathrm{~m} - 3.0 \mathrm{~m} = 2.0 \mathrm{~m} \). Use the motion equation:\[ h = \frac{1}{2}gt^2 \]\[ 2.0 \mathrm{~m} = \frac{1}{2} \times 9.81 \mathrm{~m/s^2} \times t^2 \]\[ t^2 = \frac{2 \times 2.0}{9.81} \approx 0.407 \]\[ t = \sqrt{0.407} \approx 0.64 \mathrm{~s} \]

Step 4: Determine Horizontal Distance

Using the horizontal component of velocity:\[ v_x = v\cos(\theta) = (0.1 \times 62.79)\cos(30^{\circ}) \]\[ v_x = 6.279 \times 0.866 = 5.44 \mathrm{~m/s} \]Determine the horizontal range:\[ x = v_xt \]\[ x = 5.44 \mathrm{~m/s} \times 0.64 \mathrm{~s} \approx 3.48 \mathrm{~m} \]

Key Concepts

Conservation of Energy

Conservation of energy is a fundamental principle in physics, important for understanding rotational motion. When an object rolls down an inclined plane without slipping, its mechanical energy is conserved.
This means that the initial potential energy due to height is converted into kinetic energy as it moves.
For a cylinder rolling down a slope, the potential energy (\[PE = mgh\]) initially stored due to its height off the ground is transformed into two types of kinetic energy:

• Translational kinetic energy, associated with the cylinder's movement across the plane.
• Rotational kinetic energy, connected to its spinning motion.

When the entire potential energy turns into kinetic forms, as calculated by\[mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\], this balances the translation and rotation, crucial for correctly determining the cylinder’s speed and angular velocity at the bottom of the ramp.

Moment of Inertia

The moment of inertia, often referred to as the rotational inertia, gauges how much an object resists rotational acceleration. It depends on the mass distribution relative to the axis of rotation.
The formula for a solid cylinder is given by\[I = \frac{1}{2}mr^2\],where:

• \(m\) is the mass of the cylinder.
• \(r\) is the radius of the cylinder.

Consider it as the rotational equivalent of mass in linear motion, indicating how difficult it is to change the cylinder's spinning. In exercises where energy conservation is employed, the moment of inertia is critical as it directly impacts the rotational kinetic energy involved.

Angular Velocity

Angular velocity (\(\omega\)) represents how fast an object rotates around an axis. It’s akin to linear velocity, but instead of distance per time, it involves angular displacement per time, specified typically in radians per second.
Since angular and linear velocities are linked via the equation\[v = r\omega\],knowing one gives information about the other, given the object’s radius. In problems involving cylinders, determining angular velocity is crucial for calculating the rotational kinetic energy and ensuring energy conservation between potential and kinetic states is maintained. Solving for angular velocity, particularly in cases involving rolling objects like cylinders, requires integrating the dynamics of both linear and rotational movements.

Kinematics

Kinematics involves the study of motion without addressing the forces causing it. Understanding kinematics is essential when analyzing objects on inclined planes as it helps predict the motion.
Here, equations are centered around:

• Vertical motion characterized under constant acceleration due to gravity (\(g\)).
• Horizontal motion, often uniform when considering no friction post-launch.

Analyzing the cylinder's fall from the edge, we calculated the time of flight using\[h = \frac{1}{2}gt^2\], which only relies on vertical motion parameters.
This is combined with information about horizontal velocity to calculate how far from the roof's edge the cylinder lands.

Inclined Plane

Inclined planes present scenarios where gravity’s effects are divided into components parallel and perpendicular to the plane surface.
When a cylinder rolls down such a plane, the gravitational force acts directly along the plane's angle, influencing the cylinder differently than in free falls.
Important considerations include:

• The plane's angle (\(\theta\)) alters the gravitational influence, dictating the effective height the object travels.
• Friction, in terms of rolling without slipping, is critical to ensure energy conservation laws apply correctly, as slipping would introduce additional dynamics.

This environment is instrumental in physics problems, where computation of height and potential energy dependent on angle becomes key to understanding overall motion dynamics and ensuing energy transformations.