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Determine the differential equation of motion of the 3 -kg spool. Assume that it does not slip at the surface of contact as it oscillates. The radius of gyration of the spool about its center of mass is \(k_{G}=125 \mathrm{mm}\)

Short Answer

Expert verified
The differential equation for the motion of the spool is \((\frac{a}{R^2}) = \alpha\).

Step by step solution

01

Identify the forces acting on the spool

Firstly, there are three forces acting on the spool: the gravitational force (\(F_g=mg\)), the normal force (\(F_n\)), and static friction at the point of contact (\(F_f\)). Since the spool does not slip, the friction force provides the torque that causes the spool to rotate.
02

Apply Newton's second law in the vertical direction

In the vertical direction, two forces act on the spool - the gravitational force pulling it downwards and the normal force acting upwards from the contact surface. The spool does not move vertically, so the net force in the vertical direction is zero. This gives the equation \(F_n - F_g = 0\), or \(F_n = F_g=mg\).
03

Apply Newton's second law in the horizontal direction

In the horizontal direction, there's rotational motion due to the friction force. This can be noted as \(F_f=ma\). However, since the body is rotating, we have \(a=\alpha R\), where \(\alpha\) is the angular acceleration and \(R\) the radius of the spool. Substituting, we have \(F_f=m\alpha R\).
04

Calculate the moment of inertia

The moment of inertia \(I\) of the spool about its center of mass can be calculated as \(I = mk_G^2\), where \(m\) is the mass of the spool and \(k_G\) is the radius of gyration.
05

Apply the angular version of Newton's second law

Applying the angular version of Newton's second law (\(I\alpha = Ï„\)), the friction force at the contact point causes a torque around the center. We look at both sides of the equation. On the left, \(I\alpha = mk_G^2(\frac{a}{R})\), as \(a = \alpha R\). On the right, the torque caused by friction, is \(Ï„ = F_fR = m\alpha R^2\).
06

Set up and solve the differential equation

Setting the two expressions equal gives us \(mk_G^2(\frac{a}{R}) = m\alpha R^2\). Canceling the mass, we obtain \(k_G^2(\frac{a}{R^2}) = \alpha R^2\). Hence, the differential equation for the motion of the spool is \((\frac{a}{R^2}) = \alpha\).

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Key Concepts

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

Differential Equation of Motion
When creating a differential equation of motion for an object like a spool, we begin by considering the various forces and torques acting on it. A differential equation of motion helps describe how these forces influence the spool's movement over time. In this exercise, the spool experiences no slipping at its contact surface, which simplifies the equations because frictional force exactly matches the opposing motion forces.

In particular, the derivation of the differential equation involves looking closely at both linear and angular motion. To begin, the key step is recognizing that for a stable and consistent motion, these components must balance appropriately- inertia and friction must all work cohesively. This balance leads us to formulate the linear and angular forms of Newton's Second Law, thereby creating the framework for the differential equation. This equation derived from balancing forces, moments, and accelerations unveils the mathematical relationship between the spool's movement and the forces acting on it.
Moment of Inertia
The concept of moment of inertia (I) plays an influential role in the spool's rotational dynamics. Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is akin to mass in linear motion but for rotation. For the spool, we compute moment of inertia using the formula: \[ I = mk_G^2 \] where \( m \) is the mass of the spool and \( k_G \) is the radius of gyration.

The radius of gyration (\( k_G \)) simplifies computing I as it reflects an equivalent radius at which the mass of a body could be concentrated without altering its rotational characteristics. Calculating I is crucial for understanding how much torque is needed to achieve a desired angular acceleration for the spool. It reflects how the mass is distributed with regard to its rotational axis, impacting motion predictably.
Angular Acceleration
Angular acceleration (\( \alpha \)) represents how quickly the spool's rotational speed changes. It is the rotational equivalent of linear acceleration and is related to torque and moment of inertia through Newton's laws.

The relationship is expressed through the equation \( I \alpha = \tau \), where \( \tau \) is torque. In the context of the spool problem, torque results from frictional force at its contact point causing a change in angular velocity. The formula connects the dots between linear and rotational dynamics by incorporating the spool's moment of inertia and the changing rate of its rotational speed.

Understanding \( \alpha \) helps grasp how forces result in the rotation and speed change over time and ultimately influence the spool's motion path.
Newton's Second Law
Newton's Second Law is pivotal in mechanics, outlining how the velocity of an object changes when subjected to external forces. The core of this law can be seen in its classic form: \( F = ma \), where force equals mass times acceleration.

For rotational motion, a similar principle applies where \( \tau = I \alpha \), demonstrating the balance between applied torques, moment of inertia, and angular acceleration. In the spool scenario, the second law is applied in both the vertical, horizontal, and rotational arenas, showcasing how each force component and rotational element interplays to dictate motion. It lays the foundational understanding of translating forces and torques into mathematical equations that predict an object's future position and speed.

In both linear and angular forms, Newton's Second Law connects real-world forces to dynamic motions, essential for solving problems involving movement like the spool's oscillation.

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Most popular questions from this chapter

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