Question

A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s^-1. A bob is suspended from the roof of the car by a light wire of length 1.0 m. The angle made by the wire with the vertical is (in radian)

• A. 0
• B. π/6
• C. π/3
• D. π/4

Answer 1

The Correct Option is D

Forces on the bob:

• The tension force \( T \) acts towards the center of the circle.
• The gravitational force \( mg \) acts vertically downward.

Since the bob is moving with constant speed, the bob is in equilibrium, and the net force acting on it must be zero. Now, we consider the forces in both directions:

For the vertical direction:

\(T \cos(\theta) - mg = 0\)

For the horizontal direction:

\( T \sin(\theta) = \frac{mv^2}{r} \)

From the vertical direction equation:

\( T \cos(\theta) = mg \quad \Rightarrow \quad T = \frac{mg}{\cos(\theta)} \)

Substitute this value of \( T \) into the horizontal direction equation:

\[ \frac{mg}{\cos(\theta)} \cdot \sin(\theta) = \frac{mv^2}{r} \]

Simplifying:

\[ \tan(\theta) = \frac{v^2}{rg} \]

Given values:

• Speed \( v = 10 \, \text{m/s} \)
• Radius \( r = 10 \, \text{m} \)
• Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)

Substitute these values into the equation:

\[ \tan(\theta) = \frac{10^2}{10 \times 10} = 1 \]

Now, to find the angle \( \theta \), we take the inverse tangent (arctan) of both sides:

\[ \theta = \arctan(1) = \frac{\pi}{4} \]

Therefore, the angle made by the wire with the vertical is: \( \frac{\pi}{4} \) radians (option D).

Answer 2

Let \( \theta \) be the angle made by the wire with the vertical.

We know that: 

\( \tan \theta = \frac{v^2}{r g} \)

Given: \( v = 10 \, \text{m/s} \), \( r = 10 \, \text{m} \), and \( g = 10 \, \text{m/s}^2 \), we substitute these values into the equation:

\( \tan \theta = \frac{(10 \, \text{m/s})^2}{10 \, \text{m} \times 10 \, \text{m/s}^2} = 1 \)

Now, solving for \( \theta \):

\( \theta = \tan^{-1}(1) = \frac{\pi}{4} \)

Therefore, the angle \( \theta \) is: \( \frac{\pi}{4} \).