Problem 44 - Master Resource Book in JEE Main Physics

Question

A point \(P\) moves in counter-clockwise direction on a circular path as shown in figure. The movement of \(P\) is such that it sweep out a length \(s=t^{3}+5\), where \(s\) is in metres and \(t\) is in seconds. The radius of the path is \(20 \mathrm{~m}\). The acceleration of \(P\) when \(t=2 \mathrm{~s}\) is nearly [AIEEE 2010] (a) \(14 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(13 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(12 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(7.2 \mathrm{~m} / \mathrm{s}^{2}\)

Step-by-Step Solution

Answer

The acceleration is approximately \(14 \text{ m/s}^2\), so the answer is (a).

Step 1: Understand the Problem

We need to determine the acceleration of a point moving on a circular path with radius 20 m. The path is defined by the arc length swept, given by the equation \(s = t^3 + 5\). We are asked to find acceleration at \(t = 2\) seconds.

Step 2: Differentiation to Find Velocity

First, find the linear velocity \(v\) of the point. Linear velocity is the derivative of the length function \(s\) with respect to time \(t\): \[ v = \frac{ds}{dt} = \frac{d}{dt}(t^3 + 5) = 3t^2.\] For \(t = 2\), the velocity is \[ v = 3(2)^2 = 12 \text{ m/s}.\]

Step 3: Differentiation to Find Tangential Acceleration

Determine the tangential acceleration \(a_t\), which is the second derivative of \(s\) with respect to time \(t\):\[ a_t = \frac{d^2s}{dt^2} = \frac{d}{dt}(3t^2) = 6t.\] At \(t = 2\), the tangential acceleration is \[ a_t = 6(2) = 12 \text{ m/s}^2.\]

Step 4: Calculate the Centripetal Acceleration

Centripetal acceleration \(a_c\) is given by \[ a_c = \frac{v^2}{R},\] where \(R\) is the radius of the circle. For \(v = 12 \text{ m/s}\) and \(R = 20 \text{ m}\), \[ a_c = \frac{12^2}{20} = \frac{144}{20} = 7.2 \text{ m/s}^2.\]

Step 5: Combine Accelerations

The total acceleration \(a\) is the vector sum of tangential and centripetal accelerations, computed using the Pythagorean theorem:\[ a = \sqrt{a_t^2 + a_c^2} = \sqrt{12^2 + 7.2^2} = \sqrt{144 + 51.84} = \sqrt{195.84}.\] Calculating this gives \[ a \approx 14 \text{ m/s}^2.\]

Key Concepts

Tangential Acceleration

When an object moves along a circular path, it can have tangential acceleration. This type of acceleration is responsible for changing the speed of the object as it moves along its path.
In our problem, the point moves along a circular path, where the arc length it sweeps is given by the formula \( s = t^3 + 5 \).

• The tangential acceleration \( a_t \) is found by taking the second derivative of the arc length \( s \) with respect to time \( t \).
• This is because the first derivative represents the velocity, and the second derivative gives us the change in velocity, which is acceleration.

Working through the example, we take the first derivative to find the velocity \( v = \frac{ds}{dt} = 3t^2 \). Then, take the second derivative to find the tangential acceleration:\[ a_t = \frac{d^2s}{dt^2} = 6t. \]At \( t = 2 \), substitute back to find \( a_t = 6(2) = 12 \, \text{m/s}^2 \).
This tangential acceleration indicates how quickly the speed along the path is changing.

Centripetal Acceleration

Centripetal acceleration occurs when an object is moving in a circle, constantly changing direction, even if its speed remains constant.
This acceleration always points towards the center of the circle:

• It's what keeps the object moving in its circular path.
• It is perpendicular to tangential velocity.
• The formula for centripetal acceleration is \( a_c = \frac{v^2}{R} \), where \( v \) is the linear velocity and \( R \) is the radius of the circular path.

In the given problem, the point has a velocity \( v = 12 \, \text{m/s} \) at \( t = 2 \) seconds, and the circle's radius is 20 meters.
Plugging these values into the formula, we get \[ a_c = \frac{12^2}{20} = \frac{144}{20} = 7.2 \, \text{m/s}^2. \]This centripetal acceleration ensures that the point stays moving in the circle without flying off in a straight line.

Derivatives in Physics

Derivatives in physics are an essential tool to understand how certain quantities change over time. When an equation describes a situation like an object's path or velocity, derivatives can help find how these quantities change.
Here’s why they are crucial in our problem:

• The first derivative of the position function \( s(t) = t^3 + 5 \) gives us the velocity \( v \). This shows us the speed of the point along the circular path.
• The second derivative gives the tangential acceleration \( a_t \). This shows how the speed of the point changes as it moves.
• The relationships captured by these derivatives help us understand the motion dynamics of an object experiencing both tangential and centripetal effects.

By computing derivatives, we're effectively capturing snapshots of change at each moment, allowing for precise predictions of motion characteristics like speed and acceleration. Thus, through derivatives, we gain insight into both the rate and type of motion—whether straightforward or more complex like circular motion.