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Chapter 9: Problem 4

Which of the following is incorrect about non-uniform circular motion \(\left(a_{c}=\right.\) centripetal acceleration, \(a_{t}=\) tangential acceleration)? (1) \(\vec{v} \cdot \vec{\omega}=0\) (2) \(\vec{a}_{c} \cdot \vec{v}=0\) (3) \(\vec{\omega} \cdot \vec{a}_{c}=0\) (4) \(\vec{v} \cdot \vec{a}_{t}=0\)

Short Answer

Expert verified
Option 1 is incorrect because \\(\vec{v}\\) and \\(\vec{\omega}\\) are not generally perpendicular.

Step by step solution

01

Understanding Non-uniform Circular Motion

In non-uniform circular motion, an object moves in a circular path with varying speed. It involves both centripetal (radial) acceleration, \(a_c\), directed towards the center of the circle, and tangential acceleration, \(a_t\), which changes the speed of the object along the path.
02

Analyzing Option 1

Option 1 states \( \vec{v} \cdot \vec{\omega} = 0\). The velocity vector \(\vec{v}\) is tangential to the circular path, and \(\vec{\omega}\) (angular velocity vector) is directed along the axis of rotation. These two vectors are neither parallel nor perpendicular, thus their dot product is not necessarily zero in general.
03

Analyzing Option 2

Option 2 states \( \vec{a}_c \cdot \vec{v} = 0\). Centripetal acceleration \(\vec{a}_c\) and velocity \(\vec{v}\) are perpendicular. Hence, their dot product is zero.
04

Analyzing Option 3

Option 3 states \(\vec{\omega} \cdot \vec{a}_c = 0\). \(\vec{\omega}\) is along the axis of rotation, and \(\vec{a}_c\) is radial. Since these vectors are perpendicular, their dot product is zero.
05

Analyzing Option 4

Option 4 states \( \vec{v} \cdot \vec{a}_t = 0\). In non-uniform circular motion, tangential acceleration \(\vec{a}_t\) is in the same direction as the velocity \(\vec{v}\), thus their dot product is not zero.
06

Identifying the Incorrect Statement

Compare the analyses of all options. Only Option 1 has been identified as having a dot product that is not zero because \(\vec{v}\) and \(\vec{\omega}\) are neither perpendicular nor parallel in general.

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Key Concepts

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

Centripetal Acceleration
In non-uniform circular motion, centripetal acceleration plays a vital role in guiding an object around a circular path. The centripetal acceleration, denoted as \(a_c\), is always directed towards the center of the circle.

This radial force is what keeps the object moving in a circle rather than flying off in a straight line. It changes the direction of the object鈥檚 velocity without altering its speed.

A critical point to understand is that centripetal acceleration is perpendicular to the velocity of the object as it travels along the path. This means that the centripetal acceleration does not do work on the object or change its speed, only its direction. This perpendicular relationship is also why their dot product \(\vec{a}_c \cdot \vec{v} = 0\). Remember:
  • Centripetal force keeps the object in circular motion.
  • Perpendicular to velocity, so no work is done changing speed.
Tangential Acceleration
In cases of non-uniform circular motion, tangential acceleration \(a_t\) becomes significant. Tangential acceleration is responsible for variations in the speed of the object as it moves along its circular path.

Unlike centripetal acceleration, tangential acceleration is not directed towards the center but is instead always aligned with the velocity vector. It is the reason why an object speeds up or slows down during circular motion.

The directionality of tangential acceleration, being the same as that of velocity, means that its dot product with the velocity \(\vec{v} \cdot \vec{a}_t\) is non-zero. This distinction is essential:
  • Increases or decreases speed of the object.
  • Aligned with velocity, leading to a non-zero dot product.
Dot Product Analysis
The dot product is a mathematical operation that reveals the degree of alignment between two vectors. In analyzing non-uniform circular motion, it helps discern relationships between variables like velocity \(\vec{v}\), angular velocity \(\vec{\omega}\), centripetal acceleration \(\vec{a}_c\), and tangential acceleration \(\vec{a}_t\).

To interpret the dot product results:
  • \(\vec{v} \cdot \vec{\omega}\) is typically not zero, since these vectors are not necessarily perpendicular or parallel.
  • \(\vec{a}_c \cdot \vec{v} = 0\) indicates perpendicular vectors.
  • \(\vec{\omega} \cdot \vec{a}_c = 0\) shows non-alignment as they are perpendicular.
  • \(\vec{v} \cdot \vec{a}_t\) is not zero because speed changes are in the same direction.
Understanding these relationships helps in determining incorrect assumptions about circular motion, as is illustrated when analyzing given options for correctness.

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Most popular questions from this chapter

Two particles start moving on circles of radius \(R_{1}=1 \mathrm{~m}\) and \(R_{2}=3 \mathrm{~m}\) shown with velocity \(v_{1}=2 \mathrm{~m} / \mathrm{s}, v_{2}=15 \mathrm{~m} / \mathrm{s}\) respectively. The minimum time after which the particles will again be collinear with centre is equal to (1) \(\frac{2 \pi}{3} \mathrm{sec}\) (2) \(\frac{2 \pi}{5} \mathrm{sec}\) (3) \(2 \pi \mathrm{sec}\) (4) \(\frac{\pi}{3} \mathrm{sec}\)

If the angular frequency of the rotation of the plate is \(\omega=\sqrt{\frac{g}{2 R}}\). The friction force acting on coin is (1) \(\frac{3}{4} m g \rightarrow\) (2) \(\frac{m g}{4} \leftarrow\) (3) \(\frac{m g}{2} \leftarrow\) (4) \(\frac{m g}{2} \rightarrow\)

A block of mass \(m\) is revolving in a smooth horizontal plane with a constant speed \(v .\) If the radius of the circular path is \(R\), find the total contact force received by the block. (1) \(\frac{m v^{2}}{R}\) (2) \(m g\) (3) \(m \sqrt{\frac{v^{4}}{R^{2}}+4 g^{2}}\) (4) \(m \sqrt{\frac{v^{4}}{R^{2}}+g^{2}}\)

Two identical particics we string which passes through a hole at the center of a table of the particles is made to move in a circle on that string which passes througit One of the particles is made to move in a circle on the tabl with angular velocity \(\omega\), and the other is made to \(\mathrm{m}\), in a horizontal circle as a contact pendulum with angular velocity pendulum with angular velocity \(\omega_{2} .\) If \(l_{1}\) and \(l_{2}\) are the length of he string over and under the able, then in order that particle under the table neither moves own nor moves up, the ratio \(l_{1} / l_{2}\) is (1) \(\frac{\omega_{1}}{\omega_{2}}\) (2) \(\frac{\omega_{2}}{\omega_{1}}\) (3) \(\frac{\omega_{1}^{2}}{\omega_{2}^{2}}\) (4) \(\frac{\omega_{2}^{2}}{\omega_{1}^{2}}\)

A small sphere is given vertical velocity of magnitude \(v_{0}=5 \mathrm{~ms}^{-1}\) and it swings in a vertical plane about the end of a massless string. The angle \(\theta\) with the vertical at which string will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere, is (1) \(\cos ^{-1}\left(\frac{2}{3}\right)\) (2) \(\cos ^{-1}\left(\frac{1}{4}\right)\) (3) \(60^{\circ}\) (4) \(30^{\circ}\)
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