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Chapter 10: Problem 47

A small ball of mass \(0.75 \mathrm{~kg}\) is attached to one end of a 1.25 -m-long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is \(30^{\circ}\) from the vertical, what is the magnitude of the gravitational torque calculated about the pivot?

Short Answer

Expert verified
The gravitational torque is approximately 4.59 N路m.

Step by step solution

01

Understand Torque

Torque (\( \tau \)) is the tendency of a force to rotate an object about an axis or pivot. The formula for torque is given by \( \tau = r \, F \, \sin(\theta) \), where \( r \) is the lever arm (distance from the pivot), \( F \) is the force applied, and \( \theta \) is the angle between the force and the lever arm.
02

Identify the Variables

In this problem, the mass of the ball \( m = 0.75 \, \text{kg} \), the length of the rod \( r = 1.25 \, \text{m} \) (also the lever arm since the rod is massless), and the angle \( \theta = 30^{\circ} \) from the vertical. The gravitational force acting on the ball is \( F = mg \), where \( g = 9.8 \, \text{m/s}^2 \).
03

Calculate the Gravitational Force

The gravitational force acting on the ball can be calculated as \( F = mg \), which is \( F = 0.75 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 7.35 \, \text{N} \).
04

Compute the Torque

Substitute the known values into the torque formula: \( \tau = r \, F \, \sin(\theta) = 1.25 \, \text{m} \times 7.35 \, \text{N} \times \sin(30^{\circ}) \). Calculate \( \sin(30^{\circ}) = 0.5 \). Thus, \( \tau = 1.25 \, \text{m} \times 7.35 \, \text{N} \times 0.5 = 4.59375 \, \text{N} \cdot \text{m} \).
05

Round the Result

Round the calculated torque to a reasonable level of precision. The torque is \( 4.59 \, \text{N} \cdot \text{m} \), which is the magnitude of the gravitational torque about the pivot.

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Key Concepts

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

Understanding Torque Calculation
Torque is a fundamental concept in physics, especially when dealing with rotational motion. It measures how much a force acting on an object causes it to rotate. Whenever you apply a force to an object at some distance from a pivot point, you create torque. The formula to calculate torque is given by \( \tau = r \ F \ \sin(\theta) \).

Let's break down each part of the formula:
  • \( \tau \) is the torque.
  • \( r \) is the lever arm, which is the perpendicular distance from the pivot to the line of action of the force.
  • \( F \) is the magnitude of the force applied.
  • \( \theta \) is the angle between the force and the lever arm.
To calculate, you multiply the force by the lever arm and then by the sine of the angle between them.
For our pendulum example, substituting the values into the torque formula allows you to find the rotational force acting about the pivot, which in this case is caused by gravity.
Pendulum Physics and Its Dynamics
The pendulum is a classic physics problem that beautifully illustrates rotational dynamics and the concept of torque. It consists of a mass attached at the end of a rod or string, which swings back and forth under the influence of gravity.

In the case of the simple pendulum described in the exercise, key aspects are:
  • The mass of the ball, which is gravitationally affected, causing it to swing.
  • The rod acts as a massless lever arm, meaning all of its mass is concentrated at the ball.
  • The gravitational force acts downward, and not directly along the lever arm, which is why the concept of torque is important.
The pendulum experiences maximal gravitational torque when it's at an angle from the vertical, as in the 30掳 scenario described. In this position, the gravitational force causes the greatest potential rotational energy. This is because the component of the gravitational force that acts perpendicular to the rod (which affects torque) is maximized when the pendulum is deviated from its equilibrium position.
Role of the Lever Arm in Torque
The lever arm is a critical element in calculating torque. It influences how much rotational force an applied force can exert on an object. In essence, it's the perpendicular distance from the axis of rotation (or pivot) to the line along which the force is acting.

For the pendulum:
  • The lever arm length is equivalent to the length of the rod since it is massless and only the ball at the end contributes to the dynamics.
  • In the formula \( \tau = r \ F \ \sin(\theta) \), "\( r \)" is this very lever arm, determining the turning effect of the force.
  • A longer lever arm means a greater potential for torque when the same force is applied, making it a vital factor in the magnitude of rotational motion.
Understanding how the lever arm works helps illuminate why and how objects rotate in response to different points of applied force, a key concept in both physics and everyday mechanical systems.

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Most popular questions from this chapter

A disk, with a radius of \(0.25 \mathrm{~m},\) is to be rotated like a merry-go- round through 800 rad, starting from rest, gaining angular speed at the constant rate \(\alpha_{1}\) through the first 400 rad and then losing angular speed at the constant rate \(-\alpha_{1}\) until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed \(400 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the least time required for the rotation? (b) What is the corresponding value of \(\alpha_{1} ?\)

Figure \(10-54\) shows a flat construction of two circular rings that have a common center and are held together by three rods of negligible mass. The construction, which is initially at rest, can rotate around the common center (like a merrygo-round), where another rod of negligible mass lies. The mass, inner radius, and outer radius of the rings are given in the following table. A tangential force of magnitude \(12.0 \mathrm{~N}\) is applied to the outer edge of the outer ring for \(0.300 \mathrm{~s}\). What is the change in the angular speed of the construction during the time interval?

A record turntable rotating at \(33 \frac{1}{3}\) rev \(/\) min slows down and stops in \(30 \mathrm{~s}\) after the motor is turned off. (a) Find its (constant) angular acceleration in revolutions per minute-squared. (b) How many revolutions does it make in this time?

A uniform cylinder of radius \(10 \mathrm{~cm}\) and mass \(20 \mathrm{~kg}\) is mounted so as to rotate freely about a horizontal axis that is parallel to and \(5.0 \mathrm{~cm}\) from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

The angular position of a point on a rotating wheel is given by \(\theta=2.0+4.0 t^{2}+2.0 t^{3},\) where \(\theta\) is in radians and \(t\) is in seconds. At \(t=0,\) what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at \(t=4.0 \mathrm{~s} ?\) (d) Calculate its angular acceleration at \(t=2.0 \mathrm{~s}\). (e) Is its angular acceleration constant?
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