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Chapter 9: Problem 71

Measuring \(I\). As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be \(0.640 \mathrm{~m}\). Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an \(8.20 \mathrm{~kg}\) block of wood from the free end of the rope, as in Fig. E9.49. You release the system from rest and find that the block descends \(12.0 \mathrm{~m}\) in \(4.00 \mathrm{~s}\). What is the moment of inertia of the wheel for this axis?

Short Answer

Expert verified
The moment of inertia for the wheel for this axis should be calculated using the above steps. Make sure to substitute the correct units throughout your calculations and at the end express your answer in \(kg*m^2\).

Step by step solution

01

Calculate the acceleration

The first step is to calculate the acceleration of the descending block. This can be done using the formula of constant acceleration: \(a = \frac{2(d-u*t)}{t^2}\), where \(d\) is the distance travelled, \(u\) is the initial speed that is 0 in this case because the system is released from rest, and \(t\) is the time taken. Substituting the given values, we find the acceleration \(a\) of the block.
02

Calculate the tension in the rope

The tension in the rope while the block is descending can be calculated using Newton's second law: \(F = ma\). We know the mass of the block and the acceleration from step 1 - substituting these values gives us the tension \(T\). Keep in mind that the force is the sum of tension and gravitational force, hence \(T = m*g - F\), where \(g\) denotes gravity.
03

Determine the radius of the wheel

The radius \(r\) of the wheel is needed for further calculations. Given the diameter, the radius can be calculated by dividing the diameter by 2.
04

Calculate the torque

Torque (\(\tau\)) can be calculated using the formula: \(\tau = r*T\), where \(r\) is the radius from step 3 and \(T\) is tension derived from step 2.
05

Determine the angular acceleration

The angular acceleration (\(\alpha\)) can be calculated using the relation between linear and angular acceleration: \(a = r*\alpha\). From the linear acceleration determined in step 1 and the radius we have from step 3, we can find \(\alpha\).
06

Calculate the moment of inertia

Finally, the moment of inertia (\(I\)) can be found using Newton's second law for rotation: \(\tau = I*\alpha\). The torque from step 4 and the angular acceleration from step 5 can be substituted into this equation to solve for \(I\).

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Key Concepts

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

Torque
Torque is a crucial concept when dealing with rotational motion. Imagine trying to open a door; the amount of effort required doesn't only depend on how hard you push, but also where you push. This rotational effect is torque, which is the tendency of a force to rotate an object about an axis, fulcrum, or pivot. It's determined by the equation:
  • \( \tau = r \times F \)
where \(\tau\) is the torque, \(r\) is the radius (distance from the axis of rotation to where the force is applied), and \(F\) is the force applied. In the case of the wheel, the force is the tension in the rope.

By calculating the torque, we can determine how effectively the rope's force is causing the wheel to turn. In our exercise, the tension rips around the wheel, influencing its rotation based on how forcefully and at what point the rope pulls it.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time. When a force acts on an object causing it to rotate, not only does torque come into play, but angular acceleration does as well. It's similar to linear acceleration in that it describes how quickly an object speeds up but in a rotational manner. The relationship between linear and angular acceleration is:
  • \( a = r \times \alpha \)
where \(a\) is the linear acceleration, \(r\) is the radius, and \(\alpha\) is the angular acceleration.

In the example of the wheel, the linear motion of the descending block is translated into rotation by the wheel. We measure the linear acceleration and use it to find angular acceleration. This allows us to link how fast the block descends to how quickly the wheel speeds up its rotational motion.
Newton's Second Law
Newton's second law is fundamental to understanding motion, whether it's linear or rotational. For linear motion, it states:
  • \( F = m \times a \)
where \(F\) is the force applied, \(m\) is the mass, and \(a\) is the acceleration.

For rotational motion, the equivalent form is:
  • \( \tau = I \times \alpha \)
where \(\tau\) is torque, \(I\) is the moment of inertia, and \(\alpha\) is angular acceleration. In our scenario with the wheel and block, Newton's second law helps in calculating the moment of inertia.

By understanding the forces at work and the resulting torques and accelerations, we can determine how resistant the wheel is to changes in its motion. This resistance, quantified as moment of inertia, tells us how much torque is needed for a desired angular acceleration.

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Most popular questions from this chapter

A thin, rectangular sheet of metal has mass \(M\) and sides of length \(a\) and \(b\). Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one corner of the sheet.

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant linear speed of \(v=1.25 \mathrm{~m} / \mathrm{s} .\) Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the \(\mathrm{CD}\) is played. (See Exercise \(9.20 .\) ) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta)=r_{0}+\beta \theta,\) where \(r_{0}\) is the radius of the spiral at \(\theta=0\) and \(\beta\) is a constant. On a \(\mathrm{CD}, r_{0}\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d \theta,\) the distance scanned along the track is \(d s=r d \theta .\) Using the above expression for \(r(\theta),\) integrate \(d s\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) since the track is scanned at a constant linear speed \(v,\) the distance \(s\) found in part (a) is equal to vi. Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta ;\) choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_{z}\) and the angular acceleration \(\alpha_{z}\) as functions of time. Is \(\alpha_{z}\) constant? (d) On a CD, the inner radius of the track is \(25.0 \mathrm{~mm}\), the track radius increases by \(1.55 \mu \mathrm{m}\) per revolution, and the playing time is \(74.0 \mathrm{~min} .\) Find \(r_{0}, \beta,\) and the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_{z}\) (in rad/s) versus \(t\) and \(\alpha_{z}\) (in rad/s \(^{2}\) ) versus \(t\) between \(t=0\) and \(t=74.0 \mathrm{~min}\)

A uniform sphere made of modeling clay has radius \(R\) and moment of inertia \(I_{1}\) for rotation about a diameter. It is flattened to a disk with the same radius \(R .\) In terms of \(I_{1},\) what is the moment of inertia of the disk for rotation about an axis that is at the center of the disk and perpendicular to its flat surface?

A wheel is rotating about an axis that is in the \(z\) -direction. The angular velocity \(\omega_{z}\) is \(-6.00 \mathrm{rad} / \mathrm{s}\) at \(t=0,\) increases linearly with time, and is \(+4.00 \mathrm{rad} / \mathrm{s}\) at \(t=7.00 \mathrm{~s}\). We have taken counterclockwise rotation to be positive. (a) Is the angular acceleration during this time interval positive or negative? (b) During what time interval is the speed of the wheel increasing? Decreasing? (c) What is the angular displacement of the wheel at \(t=7.00 \mathrm{~s}\) ?

Three small blocks, each with mass \(m\), are clamped at the ends and at the center of a rod of length \(L\) and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point onefourth of the length from one end.
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