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Chapter 6: Problem 63

Let \(\vec{F}\) be the force acting on a particle having position vector \(\vec{r}\), and \(\vec{\tau}\) be the torque of this force about the origin. Then [2003] (A) \(\vec{r} \cdot \vec{T}=0\) and \(\vec{F} \cdot \vec{\tau} \neq 0\) (B) \(\vec{r} \cdot \vec{T} \neq 0\) and \(\vec{F} \cdot \vec{\tau}=0\) (C) \(\vec{r} \cdot \vec{F} \neq 0\) and \(\vec{F} \cdot \vec{\tau} \neq 0\) (D) \(\vec{r} \cdot \vec{\tau}=0 \vec{F} \cdot \vec{\tau}=0\)

Short Answer

Expert verified
The short answer is: The correct option is (D) where \(\vec{r} \cdot \vec{\tau} = 0\) and \(\vec{F} \cdot \vec{\tau} = 0\). This is because the torque vector \(\vec{\tau}\) is orthogonal to both position vector \(\vec{r}\) and force vector \(\vec{F}\), resulting in their dot products being zero.

Step by step solution

01

Review properties of dot products and cross products

Recall that the dot product of two vectors is: \(\vec{A} \cdot \vec{B} = |A||B|\cos(\theta)\), where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\). Furthermore, the cross product of two vectors \(\vec{r}\) and \(\vec{F}\) is given by: \(\vec{\tau} = \vec{r} \times \vec{F}\), which results in a new vector \(\vec{\tau}\) that is orthogonal (i.e., perpendicular) to both \(\vec{r}\) and \(\vec{F}\). Now, let's study the given four options and calculate the dot products accordingly.
02

Investigate Option (A)

For option (A), we need to check if \(\vec{r} \cdot \vec{\tau} = 0\) and \(\vec{F} \cdot \vec{\tau} \neq 0\). Since we know that \(\vec{\tau}\) is orthogonal to both \(\vec{r}\) and \(\vec{F}\), their dot product will become zero. Therefore, \(\vec{r} \cdot \vec{\tau} = 0\). As for \(\vec{F} \cdot \vec{\tau}\), it should also be zero because the torque is orthogonal to the force vector. So option (A) is incorrect.
03

Investigate Option (B)

For option (B), we need to check if \(\vec{r} \cdot \vec{\tau} \neq 0\) and \(\vec{F} \cdot \vec{\tau} = 0\). We have already shown in step 2 that \(\vec{r} \cdot \vec{\tau} = 0\), so option (B) is incorrect as well.
04

Investigate Option (C)

For option (C), we need to check if \(\vec{r} \cdot \vec{F} \neq 0\) and \(\vec{F} \cdot \vec{\tau} \neq 0\). We know from Step 2 that \(\vec{F} \cdot \vec{\tau} = 0\), so option (C) is incorrect.
05

Investigate Option (D)

For option (D), we need to check if \(\vec{r} \cdot \vec{\tau} = 0\) and \(\vec{F} \cdot \vec{\tau} = 0\). As shown in Step 2, \(\vec{r} \cdot \vec{\tau} = 0\) and \(\vec{F} \cdot \vec{\tau} = 0\). Thus, option (D) is correct. So the answer to this problem is (D) \(\vec{r} \cdot \vec{\tau} = 0\) and \(\vec{F} \cdot \vec{\tau} = 0\).

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Key Concepts

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

Dot Product
In physics and mathematics, the dot product is an operation that takes two vectors and returns a scalar. The formula for the dot product of two vectors \( \vec{A} \) and \( \vec{B} \) is \( \vec{A} \cdot \vec{B} = |A||B|\cos(\theta) \), where \( |A| \) and \( |B| \) are the magnitudes of the vectors, and \( \theta \) is the angle between them.
The dot product measures how much one vector extends in the direction of another.
### Properties of Dot Product
  • If the dot product is zero, the vectors are orthogonal, which means they are at right angles to each other.
  • It is commutative, meaning \( \vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A} \).
  • The magnitude of a vector squared is the dot product of the vector with itself: \( \vec{A} \cdot \vec{A} = |A|^2 \).
Understanding the dot product helps in identifying relationships between vectors, especially in physics where forces and motion are considered.
Cross Product
The cross product, unlike the dot product, returns a vector rather than a scalar. It's used when we need to find a vector that is perpendicular to two given vectors. For vectors \( \vec{r} \) and \( \vec{F} \), their cross product is represented as \( \vec{\tau} = \vec{r} \times \vec{F} \). This \( \vec{\tau} \) is known as the torque vector.
### Features of Cross Product
  • The cross product results in a vector that is orthogonal to both input vectors.
  • The magnitude can be found using \(|\vec{r} \times \vec{F}| = |r||F|\sin(\theta)\), where \( \theta \) is the angle between the original vectors.
  • It is anti-commutative, meaning \( \vec{A} \times \vec{B} = - (\vec{B} \times \vec{A}) \).
In physics, especially in rotational dynamics, the cross product is essential because torque impacts how things rotate around an axis.
Orthogonal Vectors
Orthogonal vectors are vectors that meet at a right angle. In simpler terms, two vectors \( \vec{A} \) and \( \vec{B} \) are orthogonal if their dot product is zero, \( \vec{A} \cdot \vec{B} = 0 \). This property is crucial in determining perpendicularity in vectors.
### Characteristics of Orthogonal Vectors
  • They are independent; knowing one gives no information about the other.
  • In three-dimensional space, the cross product produces a vector orthogonal to both original vectors.
  • Orthogonality is a key idea in torque and force vectors, ensuring vectors like torque, derived from a cross product, point perpendicularly to the plane formed.
Understanding orthogonal vectors is fundamental in vector analysis and helps clarify relationships in physics, particularly in rotational motion and force interactions.

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Most popular questions from this chapter

A ring rolls without slipping on the ground. Its centre \(C\) moves with a constant speed \(u . P\) is any point on the ring. The speed of \(P\) with respect to the ground is \(v\). (A) \(0 \leq v \leq 2 u\) (B) \(v=u\), if \(C P\) is horizontal (C) \(v=u\), if \(C P\) makes an angle of \(30^{\circ}\) with the horizontal and \(P\) is below the horizontal level of \(C\) (D) \(v=\sqrt{2} u\), if \(C P\) is horizontal

Four point masses, each of value \(m\), are placed at the corners of a square \(\mathrm{ABCD}\) of side \(\ell\). The moment of inertia of this system about an axis passing through \(A\) and parallel to BD iis (A) \(3 m \ell^{2}\) (B) \(m \ell^{2}\) (C) \(2 m \ell^{2}\) (D) \(\sqrt{3} m \ell^{2}\)

A uniform rod of mass \(m\) and length \(\ell\) is placed in gravity-free space and linear impulse \(J\) is given to rod at a distance \(x=l / 4\) from centre and perpendicular to rod. Point \(A\) is at a distance \(l / 3\) from centre. $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-I } \\ \hline \text { (A) Speed of centre of rod is } & \text { (1) Zero } \\ \text { (B) Speed of point } A \text { is } & \text { (2) } \frac{5}{2} \frac{J}{m} \\ \text { (C) Speed of upper end of rod is } & \text { (3) } \frac{J}{m} \\ \text { (D) Speed of lower end of rod is } & \text { (4) } \frac{J}{2 m} \\ \text { (5) } \frac{J}{5 m} \\ \hline \end{array} $$

A thin circular ring of mass \(m\) and radius \(R\) is rotating about its axis with a constant angular velocity \(\omega\). Two objects each of mass \(M\) are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity of (A) \(\frac{\omega m}{m+M}\) (B) \(\frac{\omega m}{m+2 M}\) (C) \(\frac{\omega(m+2 M)}{m}\) (D) \(\frac{\omega(m-2 M)}{m+2 M}\)

Consider a two particle system with particles having masses \(m_{1}\) and \(m_{2}\). If the first particle is pushed towards the centre of mass through a distance \(d\), by what distance should the second particle be moved, so as to keep the centre of mass at the same position? (A) \(\frac{m_{1}}{m_{2}} d\) (B) \(d\) (C) \(\frac{m_{2}}{m_{1}} d\) (D) \(\frac{m_{1}}{m_{1}+m_{2}} d\)
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