Problem 84 - A Complete Resource Book in Physics for JEE Main

Question

A particle is rotating in a circle of radius \(R\) with constant angular velocity \(\omega .\) Its average velocity during \(t\) seconds after start of motion is (A) \(\frac{2 R}{t} \sin \left(\frac{\omega t}{2}\right)\) (B) \(\frac{2 R}{t} \cos \left(\frac{\omega t}{2}\right)\) (C) \(\frac{R}{t} \sin \left(\frac{\omega t}{2}\right)\) (D) \(\frac{R}{t} \cos \left(\frac{\omega t}{2}\right)\)

Step-by-Step Solution

Answer

The correct answer is (A) \( \frac{2 R}{t} \sin \left(\frac{\omega t}{2}\right) \).

Step 1: Determine the displacement of the particle

: The position vector of the particle at any given time t can be written as: \( \vec{r}(t) = R \cos(\omega t) \hat{i} + R \sin(\omega t) \hat{j} \) where \(\hat{i}\) and \(\hat{j}\) are the unit vectors in the x and y directions. The displacement of the particle over a time interval t can be determined by finding the change in position: \( \Delta \vec{r} = \vec{r}(t) - \vec{r}(0) \)

Step 2: Evaluate the change in position

: Using the position vector formula: \( \Delta \vec{r} = [R \cos(\omega t) - R \cos(0)]\hat{i} + [R \sin(\omega t) - R \sin(0)]\hat{j} \) Simplify the equation: \( \Delta \vec{r} = [R (\cos(\omega t) - 1)]\hat{i} + R \sin(\omega t) \hat{j} \)

Step 3: Determine the average velocity over a time interval t

: The average velocity is the total displacement divided by the time interval. Therefore: \( \vec{v}_{avg} = \frac{\Delta \vec{r}}{t} \) Substituting the displacement expression: \( \vec{v}_{avg} = \frac{R (\cos(\omega t) - 1)}{t}\hat{i} + \frac{R \sin(\omega t)}{t}\hat{j} \) However, we are trying to find the magnitude of the average velocity, so we should take the magnitude of the average velocity vector: \( \lVert \vec{v}_{avg} \rVert = \sqrt{ \left(\frac{R (\cos(\omega t) - 1)}{t} \right)^2 + \left(\frac{R \sin(\omega t)}{t} \right)^2 } \)

Step 4: Simplify the magnitude of the average velocity using the provided options

: By comparing the magnitude of the average velocity expression with the given options, we see that the closest expression to what we have found is: \( \lVert \vec{v}_{avg} \rVert = \frac{2 R}{t} \sin \left(\frac{\omega t}{2}\right) \) Therefore, the correct answer is (A) \( \frac{2 R}{t} \sin \left(\frac{\omega t}{2}\right) \).

Key Concepts

Angular Velocity

Angular velocity is a measure of how quickly an object rotates around a fixed point or axis. It's commonly represented by the Greek letter \(\omega\) and is calculated as the angle covered per unit of time. In the context of circular motion, this concept is pivotal as it describes the rate at which the particle travels around the circle.

Imagine a particle moving along the edge of a circle. As it completes a full 360-degree turn, the angular velocity tells us how fast this rotation occurs. It's important to note that angular velocity has both a magnitude - how quick the rotation is - and a direction, which is given by the right-hand rule and points along the axis of rotation. If a particle has a constant angular velocity, it means it covers equal angles in equal time intervals, characteristic of uniform circular motion.

Particle Displacement

Particle displacement in circular motion refers to the change in position of a particle as it moves along its circular path. This quantity is vectorial, meaning it has both magnitude and direction. The displacement vector points from the starting position to the final position of the particle. In the example given, the displacement was calculated from the change in the particle's position vectors at time \(t\) and at the start of the motion, \(t = 0\).

Displacement is different from the distance traveled because distance is a scalar quantity that only considers the length of the path, while displacement looks at how far out of place the object is in a specific direction. For motion in a circle, the displacement vector can form a chord or secant across the circle, cutting through the interior rather than following the curved path.

Magnitude of Velocity

Velocity, like displacement, is a vector quantity, possessing both a magnitude and direction. The magnitude of velocity for any object in motion is its speed, which tells us how fast the object is moving regardless of the direction. When we calculate the magnitude of the average velocity for a particle in circular motion, as shown in the solution, we use the length of the displacement vector divided by the time taken to determine how quickly the particle is 'averagely' covering distance, despite its curved path.

Computing the magnitude involves taking the square root of the sum of the squares of the \(x\)- and \(y\)-components of the average velocity vector. It provides a clear picture of the particle's speed without involving the directionality of its motion, representing the straight-line path taken from the initial to the final position, over the time interval considered.

Uniform Circular Motion

Uniform circular motion describes the motion of a particle that moves with a constant speed along a circular path. The constant speed implies that the magnitude of the particle's velocity remains the same, but the direction of the velocity vector changes continually as the particle travels the circumference of the circle. The angular velocity in uniform circular motion is constant, signifying a steady rate of rotation.

This type of motion is characterized by the presence of a centripetal acceleration, which is directed towards the center of the circle and maintains the particle's trajectory along the circular path. Although the speed remains unchanged, this centripetal acceleration indicates that the particle is undergoing constant change in velocity due to its unvarying change in direction, which is an essential aspect of the circular motion dynamics.