A wheel with radius 0.0600 m rotates about a horizontal frictionless axle at its center. The moment of inertia of the wheel about the axle is 2.50 kg · m2. The wheel is initially at rest. Then at t = 0 a force F(t) = (4.00 N/s)t is applied tangentially to the wheel and the wheel starts to rotate. Part A What is the magnitude of the force at the instant when the wheel has turned through 8.00 revolutions? Express your answer with the appropriate units. HẢ F = Value Units Submit Request Answer
Angular Momentum
The momentum of an object is given by multiplying its mass and velocity. Momentum is a property of any object that moves with mass. The only difference between angular momentum and linear momentum is that angular momentum deals with moving or spinning objects. A moving particle's linear momentum can be thought of as a measure of its linear motion. The force is proportional to the rate of change of linear momentum. Angular momentum is always directly proportional to mass. In rotational motion, the concept of angular momentum is often used. Since it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics. To understand the concept of angular momentum first we need to understand a rigid body and its movement, a position vector that is used to specify the position of particles in space. A rigid body possesses motion it may be linear or rotational. Rotational motion plays important role in angular momentum.
Moment of a Force
The idea of moments is an important concept in physics. It arises from the fact that distance often plays an important part in the interaction of, or in determining the impact of forces on bodies. Moments are often described by their order [first, second, or higher order] based on the power to which the distance has to be raised to understand the phenomenon. Of particular note are the second-order moment of mass (Moment of Inertia) and moments of force.
![### Educational Exercise: Rotational Motion Problem
**Problem Statement:**
A wheel with a radius of 0.0600 m rotates about a horizontal frictionless axle at its center. The moment of inertia of the wheel about the axle is \(2.50 \, \text{kg} \cdot \text{m}^2\). The wheel is initially at rest. Then at \(t = 0\) a force \(F(t) = (4.00 \, \text{N/s})t\) is applied tangentially to the wheel and the wheel starts to rotate.
**Question (Part A):**
What is the magnitude of the force at the instant when the wheel has turned through 8.00 revolutions?
**Instructions:**
- Express your answer with the appropriate units.
**Answer Box:**
\(F =\) [Value] [Units]
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**Explanation:**
To solve this problem, we need to determine the force magnitude at the specific instant.
1. **Wheel Radius and Inertia:**
- Radius (\(r\)) = 0.0600 m
- Moment of Inertia (\(I\)) = \(2.50 \, \text{kg} \cdot \text{m}^2\)
2. **Initial Condition:**
- Initially at rest (\(\omega_0 = 0\))
3. **Applied Tangential Force:**
- \(F(t) = (4.00 \, \text{N/s})t\)
4. **Required Information:**
- Number of revolutions (\(N\)) = 8.00
To find the angular displacement in radians (\(\theta\)), we use the relation:
\[ \theta = 2\pi \times N = 2\pi \times 8.00 \, \text{radians} \]
More steps follow based on the rotational motion equations and force calculations, which involve detailed derivations typically covered in physics courses related to rotational dynamics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19d7344f-b488-43af-b6d9-a8a348fcf65c%2F82d274b8-93db-42e7-9dcc-948fea539de0%2F86gmhch_processed.jpeg&w=3840&q=75)
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