Problem 38 - Master Resource Book in JEE Main Physics

Question

A car is moving along a circular path of radius \(500 \mathrm{~m}\) with a speed of \(30 \mathrm{~ms}^{-1}\). If at some instant, its speed increases at the rate of \(2 \mathrm{~ms}^{-1}\), then at that instant the magnitude of resultant acceleration will be [UP SEE 2009] (a) \(4.7 \mathrm{~ms}^{-2}\) (b) \(3.8 \mathrm{~ms}^{-2}\) (c) \(3 \mathrm{~ms}^{-2}\) (d) \(2.7 \mathrm{~ms}^{-2}\)

Step-by-Step Solution

Answer

The magnitude of the resultant acceleration is approximately \(2.7 \; \text{ms}^{-2}\) (option d).

Step 1: Identify the Known Values

We are given the following values:- Radius of the circular path, \( r = 500 \; \text{m} \).- Speed of the car, \( v = 30 \; \text{ms}^{-1} \).- Rate of increase in speed (tangential acceleration), \( a_t = 2 \; \text{ms}^{-2} \).

Step 2: Calculate the Centripetal Acceleration

The centripetal acceleration \( a_c \) is given by the formula:\[a_c = \frac{v^2}{r}\]Substitute the known values into the formula:\[a_c = \frac{(30)^2}{500} = \frac{900}{500} = 1.8 \; \text{ms}^{-2}\]

Step 3: Calculate the Resultant Acceleration

The resultant acceleration \( a_r \) is the vector sum of tangential acceleration \( a_t \) and centripetal acceleration \( a_c \). We can use the Pythagorean theorem since they are perpendicular:\[a_r = \sqrt{a_t^2 + a_c^2}\]Substitute the known values:\[a_r = \sqrt{(2)^2 + (1.8)^2} = \sqrt{4 + 3.24} = \sqrt{7.24} \approx 2.69 \; \text{ms}^{-2}\]

Step 4: Choose the Closest Answer Option

From the calculations, the magnitude of the resultant acceleration is approximately \(2.69 \; \text{ms}^{-2}\). Matching this with the given options, the closest option is (d) \(2.7 \; \text{ms}^{-2}\).

Key Concepts

Resultant Acceleration

In circular motion, the resultant acceleration is the combination of both tangential and centripetal accelerations. Imagine riding a merry-go-round. While moving in a circle, you feel forces pulling you inward and forward. These forces are

• centripetal, which keeps you moving in a circle
• tangential, which affects your speed along the path

To find the resultant acceleration, we consider both these accelerations as perpendicular to each other. This is why we can apply the Pythagorean theorem: \[a_r = \sqrt{a_t^2 + a_c^2}\]This formula sums the squares of the tangential and centripetal accelerations before taking the square root to find the magnitude. In our exercise, replacing with given values \[a_r = \sqrt{4 + 3.24} = \sqrt{7.24} \approx 2.69 \; \text{ms}^{-2}\].The option (d) matches our answer, which shows a good understanding of how these components work together in circular motion.

Centripetal Acceleration

Centripetal acceleration is essential in circular motion, acting as the force that keeps an object moving along a circular path. Rather than changing the speed of the object, it changes the direction, pulling it towards the center of the circle. The formula for centripetal acceleration is:\[a_c = \frac{v^2}{r}\]where:

• \(v\) is the speed of the object
• \(r\) is the radius of the circular path

Using our exercise's values \(v = 30\; \text{ms}^{-1}\) and \(r = 500\; \text{m}\), our calculation \(a_c = \frac{900}{500} = 1.8\; \text{ms}^{-2}\).This value tells us how fiercely the object is being pulled towards the center to maintain its curved path. Centripetal acceleration is a continuous force ensuring the curve of motion is sustained.

Tangential Acceleration

Tangential acceleration refers to the rate of change of speed of an object traveling along a circular path. Consider this as how quickly the car can speed up or slow down along the circle. If an object's speed increases, you have a positive tangential acceleration; if it decreases, the tangential acceleration is negative.For our problem, the car's speed rise is \(2\; \text{ms}^{-2}\). This is a straightforward notion - it simply means the car's speed increases by \(2\; \text{ms}^{-1}\) every second along the path. Unlike centripetal acceleration, which alters an object's direction to stay on a curve, tangential acceleration reflects changes in speed along this curve, making it important for understanding how an object's velocity continues to change during circular motion.