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Chapter 8: Problem 89

A uniform rod of mass \(M\) and length \(\ell\) can pivot freely (i.e., we ignore friction) about a hinge attached to a wall, as in Fig. 8-63. The rod is held horizontally and then released. At the moment of release, determine \((a)\) the angular acceleration of the rod, and \((b)\) the linear acceleration of the tip of the rod. Assume that the force of gravity acts at the center of mass of the rod, as shown. [\(Hint\): See Fig. 8-20g.]

Short Answer

Expert verified
(a) \( \alpha = \frac{3g}{2\ell} \), (b) \( a_{tip} = \frac{3g}{2} \).

Step by step solution

01

Define Moment of Inertia of the Rod

The moment of inertia of a rod about its pivot point is given by the formula \( I = \frac{1}{3}M\ell^2 \), where \( M \) is the mass of the rod and \( \ell \) is the length of the rod. This moment of inertia will be used in calculating the angular acceleration.
02

Calculate Torque due to Gravity

The torque \( \tau \) acting on the rod due to gravity can be calculated as \( \tau = Mg \cdot \frac{\ell}{2} \), where \( g \) is the acceleration due to gravity. The force \( Mg \) is acting downwards at the center of mass of the rod, and the perpendicular distance to the pivot is \( \frac{\ell}{2} \).
03

Find Angular Acceleration using Torque

Angular acceleration \( \alpha \) can be found using the equation \( \tau = I\alpha \), where \( \tau \) is the torque and \( I \) is the moment of inertia. Substituting the expressions: \( Mg \cdot \frac{\ell}{2} = \frac{1}{3}M\ell^2\alpha \). Solve for \( \alpha \): \( \alpha = \frac{3g}{2\ell} \).
04

Determine Linear Acceleration of Rod's Tip

The linear acceleration \( a_{tip} \) of the tip of the rod can be determined from the relationship \( a_{tip} = \alpha \cdot \ell \). Substituting \( \alpha = \frac{3g}{2\ell} \), we get \( a_{tip} = \frac{3g}{2} \).

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Key Concepts

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

Moment of Inertia
The moment of inertia is a crucial concept in rotational dynamics. It is often compared to mass in linear motion. For a rotating object, it quantifies how difficult it is to change the object's rotational motion around a specific axis. In simple terms:
  • The more mass an object has farther from the axis of rotation, the higher its moment of inertia.
  • It's a measure of an object's resistance to changes in rotational speed.
For a uniform rod pivoting about an end, the moment of inertia is calculated using the formula \( I = \frac{1}{3}M\ell^2 \). Here, \( M \) is the mass of the rod and \( \ell \) is the length of the rod. This relationship shows that the further the mass extends from the pivot point, the harder it becomes to rotate the rod. Understanding this concept is essential for analyzing any rotational system.
Torque
Torque can be thought of as the rotational equivalent of force. It measures how much a force acting on an object causes that object to rotate. Imagine pushing a door open: the farther you apply force from the hinges, the easier it is to swing the door. That's torque at play.
  • Torque depends on the magnitude of the force and the distance from the point of rotation (the lever arm).
  • Mathematically, it's expressed as \( \tau = F \times r \), where \( F \) is the force applied, and \( r \) is the distance from the pivot point.
In this exercise, the force of gravity applies a torque on the rod about its pivot point. The formula for this is \( \tau = Mg \cdot \frac{\ell}{2} \), where \( Mg \) is the gravitational force acting at the rod's center of mass, and \( \frac{\ell}{2} \) is the distance from the pivot to where gravity acts. Torque helps us determine the resulting rotational effect on the rod.
Linear Acceleration
Linear acceleration, in this context, refers to how the velocity of an object's tip changes as it rotates around a pivot. The tip's path is part of a circular arc, so its acceleration involves both direction and speed.
  • The linear acceleration \( a_{tip} \) of the tip is directly related to the rod's angular acceleration \( \alpha \).
  • This connection is established by the equation \( a_{tip} = \alpha \cdot \ell \), where \( \ell \) is the distance from the pivot to the point of interest鈥攖he rod's tip in this case.
From the given solution, the angular acceleration was determined to be \( \alpha = \frac{3g}{2\ell} \). Plugging this into the equation for linear acceleration, we find that \( a_{tip} = \frac{3g}{2} \), highlighting how the tip of the rod accelerates linearly due to the rod's rotation. Grasping this relationship is key to understanding the dynamics of rotating bodies.

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Most popular questions from this chapter

(II) Let us treat a helicopter rotor blade as a long thin rod, as shown in Fig. 8-49. \((a)\) If each of the three rotor helicopter blades is 3.75 m long and has a mass of 135 kg, calculate the moment of inertia of the three rotor blades about the axis of rotation. \((b)\) How much torque must the motor apply to bring the blades from rest up to a speed of 6.0 rev/s in 8.0 s?

(II) A sphere of radius \(r = 34.5 cm\) and mass \(m = 1.80 kg\) starts from rest and rolls without slipping down a 30.0\(^{\circ}\) incline that is 10.0 m long. \((a)\) Calculate its translational and rotational speeds when it reaches the bottom. \((b)\) What is the ratio of translational to rotational kinetic energy at the bottom? Avoid putting in numbers until the end so you can answer: \((c)\) do your answers in \((a)\) and \((b)\) depend on the radius of the sphere or its mass?

(II) A 61-cm-diameter wheel accelerates uniformly about its center from 120 rpm to 280 rpm in 4.0 s. Determine \((a)\) its angular acceleration, and \((b)\) the radial and tangential components of the linear acceleration of a point on the edge of the wheel 2.0 s after it has started accelerating.

On a 12.0-cm-diameter audio compact disc (CD), digital bits of information are encoded sequentially along an outward spiraling path. The spiral starts at radius \(R_1 = 2.5 cm\) and winds its way out to radius \(R_2 = 5.8 cm\). To read the digital information, a CD player rotates the CD so that the player's readout laser scans along the spiral's sequence of bits at a constant linear speed of 1.25 m/s. Thus the player must accurately adjust the rotational frequency \(f\) of the CD as the laser moves outward. Determine the values for \(f\) (in units of rpm) when the laser is located at \(R_1\) and when it is at \(R_2\).

(II) A turntable of radius \(R_1\) is turned by a circular rubber roller of radius \(R_2\) in contact with it at their outer edges. What is the ratio of their angular velocities, \(\omega_1/\omega_2\)?
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