/*! 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 76 A 1.6-kg grindstone in the shape... [FREE SOLUTION] | 91Ó°ÊÓ

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A 1.6-kg grindstone in the shape of a uniform cylinder of radius 0.20 m acquires a rotational rate of 24 rev/s from rest over a 6.0-s interval at constant angular acceleration. Calculate the torque delivered by the motor.

Short Answer

Expert verified
The torque delivered is approximately 0.804 Nm.

Step by step solution

01

Understand the Known Values

We have a grindstone with a mass of 1.6 kg and a radius of 0.20 m, which accelerates to a rotational speed of 24 revolutions per second (rev/s) from rest in 6.0 seconds. Since the angular acceleration is constant, we can use these values to calculate it.
02

Convert Rotational Speed

First, convert the final rotational speed from revolutions per second to radians per second. Since one revolution is equal to \(2\pi\) radians, the final speed \(\omega_f\) is computed as:\[ \omega_f = 24 \text{ rev/s} \times 2\pi \text{ rad/rev} = 48\pi \text{ rad/s} \]
03

Calculate Angular Acceleration

Angular acceleration \(\alpha\) can be calculated using the formula for constant angular acceleration:\[ \omega_f = \omega_i + \alpha t \]where \(\omega_i = 0\) (initial angular velocity), \(\omega_f = 48\pi\), and \(t = 6.0\) seconds. Solve for \(\alpha\):\[ 0 + \alpha \times 6.0 = 48\pi \]\[ \alpha = \frac{48\pi}{6} = 8\pi \text{ rad/s}^2 \]
04

Calculate Moment of Inertia

The moment of inertia \(I\) for a cylinder is given by:\[ I = \frac{1}{2} m r^2 \]Substitute \(m = 1.6\) kg and \(r = 0.20\) m:\[ I = \frac{1}{2} \times 1.6 \times (0.20)^2 = 0.032 \text{ kg m}^2 \]
05

Calculate Torque

Torque \(\tau\) can be calculated using the equation:\[ \tau = I \alpha \]Substitute the known values of \(I = 0.032\) kg m² and \(\alpha = 8\pi\) rad/s²:\[ \tau = 0.032 \times 8\pi = 0.256\pi \text{ Nm} \]Approximate to get \(\tau \approx 0.804 \text{ Nm} \) using \(\pi \approx 3.14\).

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Key Concepts

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

Angular Acceleration
To understand angular acceleration, think of how an object's rotational speed changes over time. Angular acceleration measures how quickly this speed (angular velocity) increases or decreases. It's similar to how we think about acceleration in a straight line, but instead, we're focused on circular motion. Imagine a spinning top; as it starts to spin faster, it gains angular acceleration. The formula for angular acceleration is \(\alpha = \frac{\Delta \omega}{t}\), where \(\Delta \omega\) is the change in angular velocity, and \(t\) is the time taken for this change.
  • A positive angular acceleration means the object is speeding up.
  • A negative angular acceleration indicates it's slowing down.
In our exercise, the grindstone starts from a stand-still and reaches its full speed, so it has a constant angular acceleration.
Moment of Inertia
The moment of inertia is akin to mass in linear motion. It measures an object's resistance to changes in its rotational motion. Imagine trying to twist a door handle; the further you are from the pivot point, the easier it becomes, as you've increased the moment of inertia. For different shapes, the moment of inertia has different formulas. In the case of a uniform cylinder like the grindstone, the moment of inertia \(I\) is calculated by the formula \(I = \frac{1}{2} m r^2\), where \(m\) is the mass, and \(r\) is the radius.
  • The bigger the object or its distance from the rotational axis, the higher the moment of inertia.
  • Using these principles, we calculated the grindstone's moment of inertia to be \(0.032 \, \text{kg m}^2\).
This provides a key part of our torque calculation.
Uniform Cylinder
A uniform cylinder is a solid object with a circular base, where its mass is evenly distributed throughout. It's a common shape in physics problems, particularly when discussing rotational motion. Why is this important? Well, knowing the object is a uniform cylinder allows us to use specific formulas for calculating its moment of inertia. Let's break down why this is significant:
  • The even mass distribution makes calculations straightforward.
  • You don't need to account for any differences in mass density across the object.
In practical terms, this means when solving our exercise, we apply the formula \(I = \frac{1}{2} m r^2\) directly, without any additional considerations for varying density, thus simplifying our problem.
Rotational Motion
Rotational motion happens when an object spins around an internal axis, much like how the Earth rotates on its axis. Objects in rotational motion have unique characteristics compared to linear motion. For instance, instead of speed and distance, we focus on angular velocity and the arc length. Rotational motion is often described using quantities like:
  • Angular velocity (\(\omega\)), which describes how fast the object rotates.
  • Angular acceleration (\(\alpha\)), detailing how quickly the rotation speeds up or slows down.
In the exercise, the grindstone's rotational motion went from rest to a high rotational speed. Using these principles, we were able to compute the torque needed to achieve such rotation, showing how interconnected these concepts are.

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Most popular questions from this chapter

A hollow cylinder (hoop) is rolling on a horizontal surface at speed \(\upsilon = 3.0 m/s\) when it reaches a 15\(^{\circ}\) incline. \((a)\) How far up the incline will it go? \((b)\) How long will it be on the incline before it arrives back at the bottom?

(II) In traveling to the Moon, astronauts aboard the \(Apollo\) spacecraft put the spacecraft into a slow rotation to distribute the Sun's energy evenly (so one side would not become too hot). At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis. Determine \((a)\) the angular acceleration, and \((b)\) the radial and tangential components of the linear acceleration of a point on the skin of the ship 6.0 min after it started this acceleration.

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.]

(II) The tires of a car make 75 revolutions as the car reduces its speed uniformly from 95 km/h to 55 km/h. The tires have a diameter of 0.80 m. \((a)\) What was the angular acceleration of the tires? If the car continues to decelerate at this rate, \((b)\) how much more time is required for it to stop, and \((c)\) how far does it go?

The Moon orbits the Earth such that the same side always faces the Earth. Determine the ratio of the Moon's spin angular momentum (about its own axis) to its orbital angular momentum. (In the latter case, treat the Moon as a particle orbiting the Earth.)

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