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A clay vase on a potter's wheel experiences an angular acceleration of 8.00 \(\mathrm{rad} / \mathrm{s}^{2}\) due to the application of a \(10.0-\mathrm{N} \cdot \mathrm{m}\) net torque. Find the total moment of inertia of the vase and potter's wheel.

Short Answer

Expert verified
The total moment of inertia is 1.25 kg⋅m².

Step by step solution

01

Understanding the Problem

We have a clay vase on a potter's wheel that is experiencing an angular acceleration. The problem provides us with the angular acceleration and the net torque applied to the wheel. We need to find the total moment of inertia of the vase and the potter's wheel combined.
02

Recall the Related Formula

The formula that connects torque (\(\tau\)) and angular acceleration (\(\alpha\)) with the moment of inertia (\(I\)) is given by:\[\tau = I \times \alpha\]This formula indicates that the net torque is the product of the moment of inertia and the angular acceleration.
03

Rearrange the Formula for Moment of Inertia

We need to find the moment of inertia \(I\). Rearrange the torque formula to solve for \(I\):\[I = \frac{\tau}{\alpha}\]This reveals that the moment of inertia can be determined by dividing the net torque by the angular acceleration.
04

Insert Given Values

Substitute the given values into the rearranged formula. We have \(\tau = 10.0 \, \mathrm{N} \cdot \mathrm{m}\) and \(\alpha = 8.00 \, \mathrm{rad/s^{2}}\):\[I = \frac{10.0}{8.00}\]
05

Calculate the Moment of Inertia

Carry out the division to find the moment of inertia:\[I = \frac{10.0}{8.00} = 1.25 \, \mathrm{kg \cdot m^2}\]Thus, the total moment of inertia of the vase and the potter's wheel is \(1.25 \, \mathrm{kg \cdot m^2}\).

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Key Concepts

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

Angular Acceleration
Angular acceleration is like a speed booster for objects that spin or rotate. It's all about how fast an object picks up rotational speed. If the object spins faster and faster, it has angular acceleration.
Think of a wheel starting from rest and spinning faster. The rate at which it speeds up is the angular acceleration. It's measured in radians per second squared (\( \mathrm{rad/s}^2 \)).
The clay vase on our potter's wheel has an angular acceleration of 8.00 \( \mathrm{rad/s}^2 \). This means it's turning faster and faster at this exact rate. Angular acceleration is crucial because it affects how quickly something can reach its full spinning speed. It transforms a motionless start into dynamic rotation!
Torque
Imagine you have to open a heavy door. You push on the handle, and the door swings open. The force you apply to the handle is what we call torque. Torque is a rotational force that causes an object to turn or rotate.
The further from the pivot point (like a hinge) you apply the force, the greater the torque. It's like using a longer wrench for more turning power.
In this problem, a net torque of 10.0 N·m is applied to the potter's wheel. This is what makes the wheel spin and the clay vase to accelerate angularly. In physics, torque (\( \tau \)) is calculated by multiplying the force applied (\( F \)) with the distance from the pivot point (\( r \)): - \( \tau = F \times r \)
Torque and angular acceleration are directly related through the moment of inertia. The more torque, the greater the angular acceleration, especially if the moment of inertia remains constant.
Potter's Wheel
A potter’s wheel is a classic example of rotational motion in everyday life. It's a flat circular disc that spins around an axis, much like a wheel on a car. Potters use it to shape clay into beautiful works of art.
The clay vase sits on top of this wheel, spinning at a speed determined by how the potter controls the wheel’s motion.
The mechanics of a potter's wheel tie directly into the physics of angular motion. When a net torque is applied, the wheel experiences angular acceleration. This means the vase on top will spin faster and faster. Both the vase and the wheel share the same rotational motion, and together, they contribute to the overall moment of inertia.
In essence, a potter's wheel serves as a practical real-world example of how rotational dynamics work, showcasing how torque, angular acceleration, and inertia come together in a tangible way.
Physics Problem Solving
Tackling physics problems involves breaking them down into straightforward steps, much like solving a puzzle. Let's imagine you're trying to solve a problem about a rotating object.First, understand what's given and what you need to find. In our exercise, the given factors are angular acceleration and torque. Our goal is to determine the moment of inertia.
Then, recall relevant formulas connecting these quantities. For angular motion, the relationship- \( \tau = I \cdot \alpha \) helps us see how torque relates to the moment of inertia and angular acceleration.
Next, rearrange the formula to solve for the unknown. In this case, we need to find \( I \) (moment of inertia), so rearrange it to:- \( I = \frac{\tau}{\alpha} \)
Finally, plug in the numbers to compute the solution. Always double-check your math to ensure everything is accurate.
Understanding these steps helps demystify physics problems, making them less intimidating and more manageable to tackle. Problem-solving is about clear thinking and precision, using mathematical relationships to reveal unknown variables.

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Most popular questions from this chapter

A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of gravity that is 1.28 m above the floor. His upper legs weigh 144 N and have a center of gravity that is 0.760 m above the floor. Finally, his lower legs and feet together weigh 87 N and have a center of gravity that is 0.250 m above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

A 9.75 -m ladder with a mass of 23.2 \(\mathrm{kg}\) lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 245 \(\mathrm{N}\) . At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.80 \(\mathrm{rad} / \mathrm{s}^{2}\) about an axis passing through the bottom end of the ladder. The ladder's center of gravity lies halfway between the top and bottom ends. (a) What is the net torque acting on the ladder? (b) What is the ladder's moment of inertia?

Two thin rectangular sheets (0.20 m 0.40 m) are identical. In the first sheet the axis of rotation lies along the 0.20-m side, and in the second it lies along the 0.40-m side. The same torque is applied to each sheet. The first sheet, starting from rest, reaches its final angular velocity in 8.0 s. How long does it take for the second sheet, starting from rest, to reach the same angular velocity?

A 15.0-m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is \(0.44 \mathrm{kg} \cdot \mathrm{m}^{2},\) and its radius is 0.160 \(\mathrm{m}\) . When the reel is turning, friction at the axle exerts a torque of magnitude 3.40 \(\mathrm{N} \cdot \mathrm{m}\) on the reel. If the hose is pulled so that the tension in it remains a constant \(25.0 \mathrm{N},\) how long does it take to completely unwind the hose from the reel? Neglect the mass and thickness of the hose on the reel, and assume that the hose unwinds without slipping.

As seen from above, a playground carousel is rotating counter-clockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. The carousel has a radius of \(1.50 \mathrm{m},\) an initial angular speed of \(3.14 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of 125 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) The mass of the person is 40.0 \(\mathrm{kg}\) . Find the final angular speed of the carousel after the person climbs aboard.

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