Problem 10 - Master Resource Book in JEE Main Physics

Question

The distance \(r\) from the origin of a particle moving in \(x y\)-plane varies with time as \(r=2 t\) and the angle made by the radius vector with positive \(x\)-axis is \(\theta=4 t\). Here, \(t\) is in second, \(r\) in metre and \(\theta\) in radian. The speed of the particle at \(t=1 \mathrm{~s}\) is (a) \(10 \mathrm{~ms}^{-1}\) (b) \(16 \mathrm{~ms}^{-1}\) (c) \(20 \mathrm{~ms}^{-1}\) (d) \(12 \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

The speed of the particle at \(t=1\text{ s}\) is approximately \(8.25\text{ m/s}\). None of the options directly match this value.

Step 1: Calculate Radial Speed

The radial component of speed is the derivative of the radius with respect to time. Given the equation \(r = 2t\), we differentiate with respect to \(t\) to find \(\frac{dr}{dt}\):\[\frac{dr}{dt} = \frac{d}{dt}(2t) = 2\]This means the radial speed \(v_r = 2 \text{ m/s}.\)

Step 2: Calculate Angular Speed

The tangential component of speed is related to the angle \(\theta = 4t\). The angular speed \(\omega\) is the derivative of \(\theta\) with respect to time:\[\omega = \frac{d\theta}{dt} = \frac{d}{dt}(4t) = 4\]This means the angular speed \(\omega = 4 \text{ rad/s}.\)

Step 3: Calculate Tangential Speed

The tangential speed can be found by multiplying the angular speed by the radius at time \(t\):\[v_t = \omega \cdot r = 4 \cdot (2t)\]At \(t = 1 \text{ s}\), \(r = 2\text{ m}\), thus:\[v_t = 4 \cdot 2 = 8 \text{ m/s}\]

Step 4: Determine Total Speed of Particle

The total speed of the particle is the resultant of the radial and tangential components of speed. Use the Pythagorean theorem to find the resultant speed:\[v = \sqrt{v_r^2 + v_t^2} = \sqrt{2^2 + 8^2}\]Calculate the above expression:\[v = \sqrt{4 + 64} = \sqrt{68} \approx 8.25 \text{ m/s}\]

Key Concepts

Radial Speed

When discussing projectile motion, understanding radial speed is essential. Radial speed is the rate at which the distance to a point (like the origin) changes over time. In simple terms, it's how fast the particle moves closer to or away from the origin.
In the problem, you observe that the distance from the origin, represented by the variable \( r \), is expressed as a function of time: \( r = 2t \).

• To find the radial speed, differentiate \( r \) with respect to time \( t \).
• The formula to use is \( \frac{dr}{dt} \).

Upon differentiation, \( \frac{dr}{dt} = 2 \), which tells us that the radial speed is a constant \( 2 \, \text{m/s} \). This implies that for every second, the particle moves 2 meters away from the origin.

Angular Speed

Angular speed gives insight into how quickly the angle -- that the radius vector makes with the positive x-axis -- is changing. This can be visualized as how fast the particle moves around the circle at any given time.
For this exercise, angle \( \theta \) is given as \( \theta = 4t \). This clearly indicates that \( \theta \) changes linearly over time. Here is what you need to do:

• To calculate angular speed \( \omega \), differentiate \( \theta \) with respect to \( t \).
• You should use the formula \( \omega = \frac{d\theta}{dt} \).

After differentiation, \( \omega = 4 \). This signifies that the particle's angle changes by 4 radians per second. A constant angular speed means the particle is rotating around the origin at a steady rate.

Tangential Speed

Tangential speed relates to how fast the particle moves along the circular path. It is directly linked to both the radius and the angular speed. Think of it as the speed at which the particle traces out its circular path.
Here's how it is derived for the particle:

• Calculate tangential speed \( v_t \) by multiplying the angular speed \( \omega \) by the radius \( r \).
• At time \( t = 1 \text{ s} \), the radius \( r = 2 \text{ m} \).

Use the formula \( v_t = \omega \cdot r \). Performing the calculation, \( v_t = 4 \times 2 = 8 \text{ m/s} \).
This value represents the speed of the particle along its circular path, influenced by both the spinning speed and the radius.