/*! 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 82 A bolt is threaded onto one end ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 82

A bolt is threaded onto one end of a thin horizontal rod, and the rod is then rotated horizontally about its other end. An engineer monitors the motion by flashing a strobe lamp onto the rod and bolt, adjusting the strobe rate until the bolt appears to be in the same eight places during each full rotation of the rod (Fig. 6-75). The strobe rate is 2000 flashes per second; the bolt has mass \(30 \mathrm{~g}\) and is at radius \(3.5 \mathrm{~cm}\). What is the magnitude of the force on the bolt from the rod?

Short Answer

Expert verified
The magnitude of the force on the bolt from the rod is approximately 809 N.

Step by step solution

01

- Understand the Scenario

A bolt is threaded onto a thin rod that rotates about its other end. The rotation appears synchronized with a strobe light flashing at 2000 times per second (2000 Hz). The bolt appears in eight positions during one full rotation.
02

- Determine the Rotational Speed

Since the bolt appears in the same eight places during each rotation, this means each full rotation corresponds to 8 flashes of the strobe. Given the strobe rate of 2000 flashes per second, we can find the rotational speed (frequency) of the rod by dividing the strobe rate by the number of positions: \ \[ \text{Rotation Frequency} = \frac{2000 \text{ flashes/s}}{8 \text{ positions}} = 250 \text{ rotations/s} \]
03

- Convert to Angular Velocity

Convert the frequency of rotation to angular velocity (\(\omega\)). Since \(1\) rotation is \(2\pi\) radians: \ \[ \omega = 250 \text{ rotations/s} \times 2\pi \text{ radians/rotation} = 500\pi \text{ radians/s} \]
04

- Calculate the Centripetal Force

Centripetal force \(F\) acting on the bolt due to the rotation can be found using the formula: \ \[ F = m r \omega^2 \] where \(m\) is the mass (0.030 kg), \(r\) is the radius (0.035 m), and \(\omega\) is the angular velocity (500\pi \text{ radians/s}).
05

- Plug in Values

Substitute the known values into the formula: \ \[ F = (0.030 \text{ kg})(0.035 \text{ m})(500\pi \text{ radians/s})^2 \]
06

- Simplify the Calculation

Simplify the equation step-by-step to find the centripetal force: \ \[ F = (0.030)(0.035)(500\pi)^2 \] \ \[ F = (0.030)(0.035)(250000\pi^2) \] \ \[ F = (0.030)(0.035)(250000)(9.87) \approx 809 \text{ N} \]

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Key Concepts

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

Rotational Motion
Rotational motion occurs when an object spins around an axis. In this exercise, the bolt experiences rotational motion as the rod rotates horizontally about its other end. The important aspects are the rod's length, the mass of the bolt, and the speed of rotation. Understanding rotational motion helps us to analyze the bolt鈥檚 behavior and the forces involved. Imagine the rod and bolt setup like a clock鈥檚 hand moving around the center; the bolt's path forms a circular trajectory. Each rotation of the rod and bolt system synchronizes with the strobe lamp, allowing us to measure the system's speed accurately.
Angular Velocity
Angular velocity, denoted by \( \omega \), describes how fast an object rotates around its axis. It is measured in radians per second (rad/s). For this problem, we know the bolt appears in eight positions during one complete rotation of the rod, and the strobe lamp flashes at 2000 Hz. This information helps us calculate the rotational frequency and subsequently the angular velocity. Given the rotation frequency is 250 rotations per second, and knowing that one complete rotation equals \(2\pi \) radians, we calculate angular velocity as follows: \[ \omega = 250 \times 2\pi \approx 500\pi \text{ radians/s} \]
Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path. It acts towards the center of the circle. For the bolt attached to the rod, centripetal force results from the rotational motion and keeps the bolt in its circular trajectory without flying off. The formula to calculate centripetal force is: \[ F = m r \omega^2 \] where \( F \) is the centripetal force, \( m \) is the mass of the bolt (0.030 kg), \( r \) is the radius (0.035 m), and \( \omega \) is the angular velocity (500\pi radians/s). Plugging in these values, we calculate the force needed to keep the bolt moving in a circle as: \[ F = (0.030)(0.035)(500\pi)^2 \approx 809 \text{ N} \]
Physics Problem-Solving
Solving physics problems often involves breaking down a complex scenario into manageable steps. Here鈥檚 an approach:
  • Carefully read and understand the problem.
  • Identify given data and what needs to be determined.
  • Use relevant formulas and understand how variables relate.
  • Perform calculations step-by-step to avoid errors.
  • Re-check the work and ensure the results make logical sense.
For this problem, understanding how to identify rotational frequency, convert it to angular velocity, and use it to find centripetal force is key. Stick to systematic steps for clarity and accuracy.
Educational Physics
Educational physics revolves around making abstract phenomena graspable through clear explanations and practical examples. The exercise involving a rotating rod and bolt enhances our understanding of rotational motion, angular velocity, and forces in a tangible context.

When learning physics:
  • Engage with real-life examples to solidify understanding.
  • Use visual aids, like diagrams and animations, to visualize concepts.
  • Break down complex problems into simpler parts.
  • Practice regularly to reinforce learning.
  • Don鈥檛 hesitate to seek help or use additional resources.
By frequently practicing problem-solving steps and applying physics principles, you build both confidence and competence in handling real-world applications.

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Most popular questions from this chapter

A block is projected up a frictionless inclined plane with initial speed \(v_{1}=3.50 \mathrm{~m} / \mathrm{s}\). The angle of incline is \(\theta=32.0^{\circ}\). (a) How far up the plane does it go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?

For sport, a \(12 \mathrm{~kg}\) armadillo runs onto a large pond of level, frictionless ice with an initial velocity of \(5.0 \mathrm{~m} / \mathrm{s}\) along the positive direction of an \(x\) axis. Take its initial position on the ice as being the origin. It slips over the ice while being pushed by a wind with a force of \(17 \mathrm{~N}\) in the positive direction of the \(y\) axis. In unit-vector notation, what are the animal's (a) velocity and (b) position vector when it has slid for \(3.0 \mathrm{~s}\) ?

During an Olympic bobsled run, the Jamaican team makes a turn of radius \(7.6 \mathrm{~m}\) at a speed of \(96.6 \mathrm{~km} / \mathrm{h}\). What is their acceleration in \(g\) -units? \((1 g\) -unit \(=\) \(\left.9.8 \mathrm{~m} / \mathrm{s}^{2} .\right)\)

You testify as an expert witness in a case involving an accident in which car \(A\) slid into the rear of car \(B\), which was stopped at a red light along a road headed down a hill (Fig. 6-74). You find that the slope of the hill is \(\theta=12.0^{\circ}\), that the cars were separated by distance \(d=24.0 \mathrm{~m}\) when the driver of car \(A\) put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car \(A\) at the onset of braking was \(v_{1}=18.0 \mathrm{~m} / \mathrm{s}\). With what speed did car \(A\) hit car \(B\) if the coefficient of kinetic friction was (a) \(0.60\) (dry road surface) and (b) \(0.10\) (road surface covered with wet leaves)?

A \(40 \mathrm{~kg}\) slab rests on a frictionless floor. A \(10 \mathrm{~kg}\) block rests on top of the slab (Fig. 6-58). The coefficient of static friction \(\mu^{\text {stat }}\) between the block and the slab is \(0.60\), whereas their kinetic friction coefficient \(\mu^{\text {kin }}\) is \(0.40\). The \(10 \mathrm{~kg}\) block is pulled by a horizontal force with a magnitude of \(100 \mathrm{~N}\). What are the resulting accelerations of (a) the block and (b) the slab?
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