/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A thin, horizontal rod with leng... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin, horizontal rod with length \(l\) and mass \(M\) pivots about a vertical axis at one end. A force with constant magnitude \(F\) is applied to the other end, causing the rod to rotate in a horizontal plane. The force is maintained perpendicular to the rod and to the axis of rotation. Calculate the magnitude of the angular acceleration of the rod.

Short Answer

Expert verified
The magnitude of the angular acceleration \(\alpha\) is \(\frac{3F}{Ml}\).

Step by step solution

01

Understanding Torque

Torque, \(\tau\), is the measure of the force causing the object to rotate. It's given by \(\tau = r \times F\), where \(r\) is the distance from the pivot point to the point of force application. In this problem, \(r = l\) (the length of the rod). Consequently, the torque applied is \(\tau = l \cdot F\).
02

Apply Newton's Second Law for Rotation

Newton's second law for rotational motion states \(\tau = I\alpha\), where \(I\) is the moment of inertia and \(\alpha\) is the angular acceleration. For a rod rotating about one end, the moment of inertia \(I\) is \(\frac{1}{3}Ml^2\).
03

Substitute and Solve for Angular Acceleration

We now substitute the values into the rotational motion equation: \(l \cdot F = \left(\frac{1}{3}Ml^2\right)\alpha\). To find \(\alpha\), we rearrange this to \(\alpha = \frac{Fl}{\frac{1}{3}Ml^2} = \frac{3F}{Ml}\).

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Key Concepts

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

Torque
Torque is a concept that helps us understand how forces cause things to rotate. Imagine pushing a door open: the further from the hinge you push, the easier it is to open. That's torque in action! It's calculated as the product of the force applied and the distance from the pivot point, given by the formula \( \tau = r \times F \), where
  • \( \tau \) is the torque,
  • \( r \) is the distance from the pivot (or hinge),
  • \( F \) is the force applied.
In this exercise, the torque on the rod is determined by multiplying the force applied by the length of the rod, since the force is applied perpendicular to the rod. This setup gives us \( \tau = l \cdot F \). By maximizing this product, you maximize the torque, making it possible to rotate the rod more efficiently.
Moment of Inertia
The moment of inertia is like the rotational version of mass. It's a measure of how difficult it is to change an object's rotational motion. Different shapes and mass distributions will lead to different moments of inertia.For the rod in this problem, which pivots about one end, the moment of inertia is calculated as \( I = \frac{1}{3}Ml^2 \). Here,
  • \( M \) is the mass of the rod,
  • \( l \) is the length of the rod.
This formula tells us that longer rods or ones with more mass have a higher moment of inertia, making them harder to spin. Understanding the moment of inertia helps us grasp why different rotational bodies react differently to forces.
Rotational Motion
Rotational motion is when an object spins around a fixed axis. Just like linear motion depends on speed and mass, rotational motion depends on angular speed and moment of inertia. When objects undergo rotational motion, they follow laws similar to those of linear motion, such as momentum and acceleration. The force causing rotation is given by torque—similar to how force causes linear acceleration. In scenarios like this rod rotation exercise, the rod rotates in a horizontal plane due to an external force. The interrelation of torque, angular acceleration, and moment of inertia explains the dynamics of such a system, allowing us to predict rotational behavior in real life.
Newton's Second Law of Rotation
Newton's Second Law of Rotation connects how torque influences rotational motion. It states that the torque applied is equal to the moment of inertia times angular acceleration: \( \tau = I\alpha \).This equation works similarly to Newton's second law for linear motion (\( F = ma \)): instead of mass, we have moment of inertia, and instead of linear acceleration, we have angular acceleration. This means:
  • \( \tau \) is the torque,
  • \( I \) is the moment of inertia,
  • \( \alpha \) is the angular acceleration.
By rearranging this equation, you can solve for the angular acceleration (\( \alpha \)) if the torque (\( \tau \)) and the moment of inertia (\( I \)) are known, as done in this exercise where \( \alpha = \frac{3F}{Ml} \). Understanding this law reveals how applying torque affects an object's rotational acceleration, impacting everything from simple machines to complex mechanical systems.

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Most popular questions from this chapter

A hollow, spherical shell with mass 2.00 \(\mathrm{kg}\) rolls without slipping down a \(38.0^{\circ}\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to 4.00 \(\mathrm{kg}\) ?

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A small 10.0-g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50.0 \(\mathrm{g}\) and is 100 \(\mathrm{cm}\) in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 \(\mathrm{cm} / \mathrm{s}\) relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps? (b) What is the total kinetic energy of the system just after the bug leaps? (c) Where does this energy come from?

Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. The clock hand has a length of 15.0 \(\mathrm{cm}\) and a mass of 6.00 \(\mathrm{g}\) . Take the second hand to be a slender rod rotating with constant angular velocity about one end.

A horizontal plywood disk with mass 7.00 \(\mathrm{kg}\) and diameter 1.00 \(\mathrm{m}\) pivots on frictionless bearings about a vertical axis through its center. You attach a circular model-railroad track of negligible mass and average diameter 0.95 \(\mathrm{m}\) to the disk. A \(1.20-\mathrm{kg}\) , battery-driven model train rests on the tracks. To demonstrate conservation of angular momentum, you switch on the train's engine. The train moves counterclockwise, soon attaining a constant speed of 0.600 \(\mathrm{m} / \mathrm{s}\) relative to the tracks. Find the magnitude and direction of the angular velocity of the disk relative to the earth.

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