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At time \(t=0\), a \(3.0 \mathrm{~kg}\) particle with velocity \(\vec{v}=(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(6.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) is at \(x=3.0 \mathrm{~m}, y=8.0 \mathrm{~m}\). It is pulled by a \(7.0 \mathrm{~N}\) force in the negative \(x\) direction. About the origin, what are (a) the particle's angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?

Short Answer

Expert verified
(a) Angular momentum: \(-174.0\hat{k}\) kg·m²/s. (b) Torque: \(-56.0\hat{k}\) N·m. (c) Rate of change: \(-56.0\) N·m.

Step by step solution

01

Find Initial Angular Momentum

Angular momentum \( \vec{L} \) with respect to a point is given by \( \vec{L} = \vec{r} \times \vec{p} \), where \( \vec{r} \) is the position vector from the origin and \( \vec{p} \) is the linear momentum \( \vec{p} = m\vec{v} \). For the particle, \( \vec{r} = 3.0 \hat{i} + 8.0 \hat{j} \) m and \( \vec{v} = 5.0 \hat{i} - 6.0 \hat{j} \) m/s. Thus, linear momentum is \( \vec{p} = 3.0 \left(5.0 \hat{i} - 6.0 \hat{j} \right) = 15.0 \hat{i} - 18.0 \hat{j} \) kg·m/s. The angular momentum is \( \vec{L} = \left(3.0 \hat{i} + 8.0 \hat{j} \right) \times \left( 15.0 \hat{i} - 18.0 \hat{j} \right) \). Using the cross product formula, \( \vec{L} = (3.0 \times -18.0 - 8.0 \times 15.0) \hat{k} = (-54.0 - 120.0) \hat{k} = -174.0 \hat{k} \) kg·m²/s.
02

Calculate the Torque

Torque \( \vec{\tau} \) is given by \( \vec{\tau} = \vec{r} \times \vec{F} \), where \( \vec{F} \) is the force acting on the particle. The force is \( 7.0 \hat{i} \), hence \( \vec{\tau} = \left(3.0 \hat{i} + 8.0 \hat{j} \right) \times -7.0 \hat{i} \). The cross product is \( \left(3.0 \right) \times \left(-7.0 \right) \) gives zero for \( i \) component and \( \left(8.0 \hat{j} \right) \times \left(-7.0 \right) \hat{i} = 8.0 \times -7.0 \hat{k} = -56.0 \hat{k} \) N·m.
03

Determine Rate of Change of Angular Momentum

The rate of change of angular momentum is equal to the torque on the object. Therefore, \( \frac{d\vec{L}}{dt} = \vec{\tau} = -56.0 \hat{k} \) N·m. Thus, the rate at which the angular momentum is changing is \(-56.0\) N·m.

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Key Concepts

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

Torque Calculation
Torque is a crucial concept in mechanics that describes how forces cause objects to rotate. It is often symbolized as \( \vec{\tau} \) and calculated using the equation \( \vec{\tau} = \vec{r} \times \vec{F} \), where \( \vec{r} \) is the position vector from the pivot point (usually the origin) to the point where the force is applied, and \( \vec{F} \) is the force vector.
When calculating the torque in the given problem, the particle experiences a force of \(-7.0 \hat{i}\) N, indicating the negative \(x\)-direction.
To solve for torque, consider:
  • The position vector \(\vec{r} = 3.0 \hat{i} + 8.0 \hat{j} \).
  • Perform the cross product: The distance multiplied by the force (\
Rate of Change of Angular Momentum
The rate of change of angular momentum provides insights into how fast the angular momentum of a system is evolving.
In rotational dynamics, Newton's Second Law translates to the rotational form, which can be represented as: \( \frac{d\vec{L}}{dt} = \vec{\tau} \), where \( \tau \) is torque.
For a constant torque, like in this problem, this rate is simply the torque value itself.
The particle's angular momentum rate of change was given to be \(-56.0 \hat{k} \) N·m, identical to the torque in this scenario.
This equality highlights a key concept in mechanics: when an object is subject to a constant external torque, its angular momentum changes at a rate equal to that torque.
This relationship emphasizes the close link between torque and changes in rotational motion, a crucial aspect of rotational dynamics.
  • Understand that this rate can vary if the torque varies over time.
  • This principle shows that applying torque has a direct and potentially altering effect on the system's rotational state.
Understanding these aspects will help build intuition into rotational dynamics.

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Most popular questions from this chapter

A uniform disk of mass \(10 m\) and radius \(3.0 r\) can rotate freely about its fixed center like a merry-go-round. A smaller uniform disk of mass \(m\) and radius \(r\) lies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity of \(20 \mathrm{rad} / \mathrm{s}\). Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterward, the two disks again rotate together (without further sliding). (a) What then is their angular velocity about the center of the larger disk? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the two-disk system to the system's initial kinetic energy?

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