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A meterstick pivots freely from one end. If it's released from a horizontal position, find its angular velocity when it passes through the vertical. Treat the stick as a uniform thin rod.

Short Answer

Expert verified
Use energy conservation principles to find angular velocity.

Step by step solution

01

Understand the Problem

We need to find the angular velocity of a uniform thin rod, hinged at one end, as it swings from a horizontal to vertical position. This involves using principles from mechanics, specifically rotational motion under the influence of gravity.
02

Determine Initial Potential Energy

When the meterstick is horizontal, it has gravitational potential energy and no kinetic energy. If we assume the meterstick is 1 meter long and mass is uniformly distributed, its center of mass is at 0.5 m height. Thus, the initial potential energy is given by \( PE = mgh \) where \( h = 0.5 \), \( m \) is mass, and \( g \approx 9.8 \).

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Key Concepts

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

Rotational Motion
Rotational motion refers to the movement of a body around a pivot or axis. In the case of the meterstick, it rotates around the point where it is hinged. As the stick swings down, its motion is analogous to what you'd observe with any object that pivots—and this is crucial in physics. Angular velocity (\(\omega\)) is a key component here. It represents how quickly the rod rotates. When released from the horizontal position, the stick accelerates due to gravity, transforming potential energy into rotational kinetic energy until it passes through the vertical. At this point, the angular velocity reaches its maximum before gravity acts to slow it again.Key components that affect rotational motion include:
  • Inertia: A measure of how easily an object rotates. For a rod, this depends on its length and distribution of mass.
  • Torque: The force that causes the rotational motion. When gravity pulls the stick down, it creates torque.
Understanding these principles helps to calculate the maximum angular velocity of the rod when it becomes vertical.
Potential Energy
Potential energy (PE) is a form of energy that is stored within a system due to its position or configuration. In the case of our meterstick, gravitational potential energy is relevant. Initially, the rod is positioned horizontally, which means that its potential energy is at its peak because the center of mass is elevated.The potential energy of the rod can be calculated using the formula:
  • \[ PE = mgh \] where:
    • \( m \) is the mass of the rod
    • \( g \approx 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity
    • \( h \) is the height of the center of mass from the pivot, which for the meterstick is 0.5 meters when horizontal
As the rod swings down, potential energy is converted into kinetic energy, causing the rod to gain speed and rotate faster. By the time the rod reaches a vertical position, all the initial potential energy is transferred into rotational kinetic energy, and this energy transformation is what propels the rod to reach its highest angular velocity at the vertical position.
Uniform Thin Rod
A uniform thin rod is an idealization used in physics to simplify the analysis of rotational dynamics. This assumption means the mass of the rod is consistently distributed along its length, leading to a straightforward calculation of its properties.For a uniform rod:
  • Center of Mass: The center of mass (COM) is at the midpoint of the rod length. Here, it's 0.5 meters from the pivot when considering a stick of 1 meter.
  • Moment of Inertia: \[ I = \frac{1}{3} mL^2 \] where \( m \) is the mass and \( L \) is the length of the rod.
    This tells us how the rod's mass distribution affects its rotational acceleration.
These characteristics help in analyzing problems involving rotational motion because they determine how gravity affects the rod's swing and how the energy transformation occurs. Understanding these properties facilitates calculating how changes in potential energy convert to rotational motion as the uniform rod pivots from horizontal to vertical.

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Most popular questions from this chapter

If the polar ice caps melt, adding more liquid water to the oceans, Earth's rotational inertia could increase by as much as an estimated \(0.3 \%\). Compute the effect such a change would have on the length of 1 day, assuming Earth's angular momentum remains constant.

While standing on the rotating Earth, is your centripetal acceleration greater on the equator or at latitude \(45^{\circ}\) north?

A \(145-\mathrm{g}\) baseball has radius \(3.7 \mathrm{~cm}\). (a) Assuming uniform density, what's its rotational inertia? (b) The ball is pitched at \(22 \mathrm{~m} / \mathrm{s}\) with spin rate \(20 \mathrm{~Hz}\). Find and compare the ball's translational and rotational kinetic energies.

A turntable with a rotational inertia \(0.225 \mathrm{~kg} \cdot \mathrm{m}^{2}\) is rotating at \(3.25 \mathrm{rad} / \mathrm{s} .\) Suddenly, a disk with rotational inertia \(0.104 \mathrm{~kg} \cdot \mathrm{m}^{2}\) is dropped onto the turntable with its center on the rotation axis. Assuming no outside forces act, what's the common rotational velocity of the turntable and disk?

Your car tire has radius \(31.0 \mathrm{~cm}\). (a) If it rolls without slipping and has angular velocity \(79.3 \mathrm{rad} / \mathrm{s}\), what's your car's speed? Now suppose your car has the speed found in part (a), but different angular velocity. Describe what's happening to the tire where it contacts the road when (b) \(\omega=91.5 \mathrm{rad} / \mathrm{s}\) and \((\mathrm{c}) \omega=\) \(52.0 \mathrm{rad} / \mathrm{s}\)

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