Problem 83 - University Physics with Modern Physics

Question

A meter stick with a mass of 0.160 \(\mathrm{kg}\) 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 \(1.00 \mathrm{m},\) starting from rest.

Step-by-Step Solution

Answer

(a) -0.784 J, (b) 5.42 rad/s, (c) 5.42 m/s, (d) Stick's end is faster at 5.42 m/s compared to 4.43 m/s.

Step 1: Calculate Initial Gravitational Potential Energy

Initially, the stick is horizontal, and we consider the center of mass to be at the midpoint of the stick, which is 0.50 m from the pivot point. The formula for gravitational potential energy (GPE) is \[\text{GPE} = mgh\]where \(m\) is the mass, \(g\) is the acceleration due to gravity \((9.8 \ \text{m/s}^2)\), and \(h\) is the height.For the horizontal position:\[\text{Initial GPE} = 0.160 \times 9.8 \times 0.50 = 0.784 \, \text{J}\]

Step 2: Determine Final Gravitational Potential Energy

As the stick reaches the vertical position, the center of mass is directly below the pivot point, meaning the height is zero. Therefore, the final gravitational potential energy is:\[\text{Final GPE} = 0 \, \text{J}\]

Step 3: Calculate the Change in Gravitational Potential Energy

The change in gravitational potential energy is the difference between the initial and final values:\[\Delta \text{GPE} = \text{Final GPE} - \text{Initial GPE} = 0 - 0.784 = -0.784 \, \text{J}\]

Step 4: Calculate Angular Speed Using Energy Conservation

The loss in gravitational potential energy is converted to rotational kinetic energy given by:\[\text{Rotational KE} = \frac{1}{2}I\omega^2\]where \(I\) is the moment of inertia \(= \frac{1}{3}mL^2\) for a rod pivoting at the end, and \(\omega\) is the angular speed.Thus,\[0.784 = \frac{1}{2} \times \left(\frac{1}{3} \times 0.160 \times 1.00^2\right) \omega^2\]Solving for \(\omega\):\[\omega = \sqrt{\frac{2 \times 0.784}{\frac{1}{3} \times 0.160}} = 5.42 \ \text{rad/s}\]

Step 5: Determine Linear Speed of the Opposite End

The linear speed \(v\) can be related to the angular speed \(\omega\) by the formula:\[v = \omega L\]where \(L = 1.00 \ \text{m}\) is the length of the stick. Substituting in the value of \(\omega\):\[v = 5.42 \times 1.00 = 5.42 \ \text{m/s}\]

Step 6: Compare with the Speed of a Particle Dropped 1 m

A particle falling, starting from rest, has a speed calculated by:\[v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.00} = 4.43 \, \text{m/s}\]The stick's end is faster because it gains rotational kinetic energy.

Key Concepts

Angular Speed

Angular speed is a measure that describes how quickly an object rotates around an axis. In this exercise, the angular speed of the meter stick as it swings through the vertical position is determined using the conservation of energy principle.

The gravitational potential energy lost by the stick is converted into rotational kinetic energy. The formula for rotational kinetic energy is\[\text{Rotational KE} = \frac{1}{2}I\omega^2\]where:

• \(I\) is the moment of inertia, which represents how the mass of an object is distributed in relation to the axis of rotation.
• \(\omega\) is the angular speed.

For a rod pivoting at the end, the moment of inertia \(I\) is \(\frac{1}{3}mL^2\). By relating the energy lost to the kinetic energy gained, you can solve for \(\omega\), which was found to be \(5.42 \ \text{rad/s}\).

Understanding angular speed is crucial in studying rotational motion because it provides insight into how fast an object spins, which is often required for dynamic analysis in physics.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. In simpler terms, it depends on how high above the ground an object is located.

Initially, when the meter stick is held horizontally, its center of mass is at a height of 0.50 meters from the pivot. The formula to find GPE is\[\text{GPE} = mgh\]where:

• \(m\) is the mass of the object (0.160 kg for the stick).
• \(g\) is the gravitational acceleration (approximately \(9.8 \ \text{m/s}^2)\).
• \(h\) is the height of the center of mass above the pivot point.

The initial GPE is calculated to be 0.784 J. When the stick swings to a vertical position, the height is essentially zero, making the final GPE zero. This change in potential energy, \(-0.784 \ \text{J}\), is exactly the amount of energy that goes into making the stick rotate.

Linear Speed

Linear speed refers to how fast a point on the rotating object is moving along its path. For the meter stick in this exercise, the linear speed at the end opposite the pivot is what we're interested in.

The linear speed \(v\) is connected to the angular speed \(\omega\) by the formula:\[v = \omega L\]where:

• \(\omega\) is the angular speed (found to be \(5.42 \ \text{rad/s})\).
• \(L\) is the length of the stick (1.00 m).

By substituting \(\omega = 5.42 \ \text{rad/s}\) and \(L = 1.00 \ \text{m}\), we find that the end of the stick moves with a linear speed of \(5.42 \ \text{m/s}\).

Linear speed is significant in physics because it helps describe the actual motion as observed from a non-rotating perspective, allowing us to understand how quickly an object moves from one point to another.

Moment of Inertia

Moment of inertia is a property that measures how much torque is needed for a specific angular acceleration of an object. Put simply, it tells us how hard it is to spin an object.

For a rod pivoting at one end, like the meter stick in this exercise, the moment of inertia \(I\) can be calculated using the formula:\[I = \frac{1}{3}mL^2\]where:

• \(m\) is the mass of the rod.
• \(L\) is its length.

In our case, \(m = 0.160 \ \text{kg}\) and \(L = 1.00 \ \text{m}\), so the moment of inertia helps determine how the gravitational potential energy converts into rotational kinetic energy.

Understanding the moment of inertia is essential for studying rotational motion because it affects how objects respond to torques and change in angular motion. It’s the rotational analogue to mass in linear motion.