Problem 71 - University Physics with Modern Physics

Question

A meter stick with a mass of 0.180 kg is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released. As it swings through the vertical, calculate (a) the change in gravitational potential energy that has occurred; (b) the angular speed of the stick; (c) the linear speed of the end of the stick opposite the axis. (d) Compare the answer in part (c) to the speed of a particle that has fallen 1.00 m, starting from rest.

Step-by-Step Solution

Answer

Change in potential energy is 0.8829 J; angular speed is 5.42 rad/s; linear speed is 5.42 m/s; stick is faster by 0.99 m/s.

Step 1: Calculate the Change in Gravitational Potential Energy

The change in gravitational potential energy can be calculated using the formula \( \Delta U = mgh \). Here, \( m = 0.180 \) kg is the mass of the stick, \( g = 9.81 \) m/s² is the acceleration due to gravity, and \( h = \frac{L}{2} = 0.5 \) m is the change in height of the center of mass as the meter stick swings from horizontal to vertical. Therefore, \( \Delta U = 0.180 \times 9.81 \times 0.5 = 0.8829 \) J.

Step 2: Determine the Angular Speed of the Stick

Using energy conservation, the change in gravitational potential energy equals the rotational kinetic energy at the bottom, given by \( \Delta U = \frac{1}{2} I \omega^2 \). The rotational inertia of a rod pivoted about one end is \( I = \frac{1}{3} mL^2 \). Substitute \( I = \frac{1}{3} \times 0.180 \times 1^2 = 0.060 \) kg·m² into the equation: \( 0.8829 = \frac{1}{2} \times 0.060 \times \omega^2 \). Solving for \( \omega \), we find \( \omega = 5.42 \) rad/s.

Step 3: Compute the Linear Speed of the Opposite End of the Stick

The linear speed \( v \) of the end of the stick can be found from \( v = \omega L \), where \( L = 1 \) m. Substituting \( \omega = 5.42 \) rad/s gives \( v = 5.42 \times 1 = 5.42 \) m/s.

Step 4: Compare with a Particle Falling 1.00 m

For a particle dropping 1.00 m, starting from rest, use the formula \( v = \sqrt{2gh} \) to find its final speed. Substituting \( g = 9.81 \) m/s² and \( h = 1.00 \) m, we get \( v = \sqrt{2 \times 9.81 \times 1} = 4.43 \) m/s. Thus, the speed of the end of the stick is faster than that of a falling particle by approximately 0.99 m/s.

Key Concepts

Gravitational Potential Energy

Gravitational potential energy is like a stored treasure of energy due to an object's height and mass. Imagine you have a meter stick held horizontally at a pivot point. Initially, it doesn’t seem to have much energy in terms of movement. However, because of its position, it has a certain amount of energy stored due to gravity pulling on it. When the stick is released and swings down to a vertical position, the gravitational potential energy changes. We can calculate this change using the formula: \[ \Delta U = mgh \]Where:

• \( m \) is the mass of the stick (0.180 kg in our case),
• \( g \) is the acceleration due to gravity (9.81 m/s²),
• \( h \) is the change in height of the center of mass.

For our meter stick, since it's pivoted at one end, the center of mass falls by half the meter stick's length (0.5 m). So, the gravitational potential energy decreases by 0.8829 joules as the stick swings down.

Angular Speed

Angular speed describes how quickly an object is rotating. For the meter stick, once it's released, it accelerates as it swings down. By the time it reaches the vertical position, it’s rotating at its maximum speed.Energy conservation helps us find this speed. As the stick swings, the gravitational potential energy lost transforms into rotational kinetic energy, given by: \[ \Delta U = \frac{1}{2} I \omega^2 \]Where:

• \( I \) is the rotational inertia, which is \( \frac{1}{3} \times m \times L^2 \)
• \( \omega \) is the angular speed.

For this stick, \( I = 0.060 \) kg·m². By setting \( \Delta U = 0.8829 \) J equal to \( \frac{1}{2} I \omega^2 \) and solving for \( \omega \), we find the angular speed to be approximately 5.42 rad/s.

Linear Speed

Linear speed is about how fast something is moving along a path, like a car on a highway. For the swinging meter stick, we often want to know how fast the end of the stick is moving as it swings. We can find this using the relationship between angular speed and linear speed: \[ v = \omega L \]Where:

• \( v \) is the linear speed,
• \( \omega \) is the angular speed (5.42 rad/s),
• \( L \) is the length of the stick (1 m).

Plugging these values in, the linear speed of the end of the stick is 5.42 m/s. It’s like imagining the tip of the stick racing through space as it pivots.

Energy Conservation

Energy conservation is the principle that energy cannot be created or destroyed—only transformed from one form to another.
In the context of our meter stick swinging, as it falls, what happens is a beautiful dance of energy types. The gravitational potential energy decreases, but it doesn't just vanish. Instead, it transforms into rotational kinetic energy, which is why the stick speeds up as it swings down. By understanding this transformation, we can predict how fast objects will move under gravity's influence. Energy conservation lets us understand why we can use the potential energy at the top to find the rotational speed at the bottom. It links initial conditions with movement, allowing us to solve many real-world problems with physics!

Rotational Inertia

Rotational inertia, also known as the moment of inertia, tells us how much an object resists changes to its rotational motion. It's similar to mass in linear motion—more inertia means it's harder to spin something or stop it from spinning.For a meter stick pivoted at an end, the rotational inertia is calculated using the formula: \[ I = \frac{1}{3} mL^2 \]Here, each variable tells us:

• \( m \) is the mass of the stick,
• \( L \) is the length of the stick.

In our exercise, \( I = 0.060 \) kg·m². This means the stick has a small tendency to resist rotational changes because it's relatively light and not very long. Understanding rotational inertia helps us grasp why some objects spin faster or slower when influenced by forces.