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Chapter 4: Problem 44

A plumb line is suspended from the roof of a railroad car and the car is moving on a circular road. The inclination of the string of the plumb line w.r. vertical direction is (a) \(\theta=\tan ^{-1}\left(v^{2} r g\right)\) (b) \(\theta=\tan ^{-1}\left(r g / v^{2}\right)\) (c) \(\theta=\tan ^{-1}\left(v^{2} / r g\right)\) (d) \(\theta=\operatorname{lan}^{-1}\left(v^{2} r / g\right)\)

Short Answer

Expert verified
The correct answer is (c) \( \theta = \tan^{-1}\left(\frac{v^2}{rg}\right) \).

Step by step solution

01

Understanding the Forces

To solve this problem, we need to consider the forces acting on the plumb line as the car moves in a circular motion. The forces involved are the gravitational force acting vertically downward and the centripetal force acting horizontally outwards due to the circular motion.
02

Identifying the Inclination

The inclination of the string, which is represented by the angle \( \theta \), is the result of the balance between the horizontal centripetal force and the vertical weight of the plumb bob. The angle changes with the ratio of these forces.
03

Relating Forces to Angle

The centripetal force can be expressed as \( F_c = m \frac{v^2}{r} \) where \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius of the circle. The gravitational force is \( F_g = mg \). The tangent of the angle \( \theta \) is given by the ratio \( \tan \theta = \frac{F_c}{F_g} = \frac{v^2/r}{g} \) which simplifies to \( \tan \theta = \frac{v^2}{rg} \).
04

Determine the Correct Option

Given the options provided, we compare the expression \( \tan \theta = \frac{v^2}{rg} \) to the solutions. Here, option (c) fits the derived expression: \( \theta = \tan^{-1}\left(\frac{v^2}{rg}\right) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
In the world of circular motion, centripetal force plays a vital role. It is the force that keeps an object moving along a curved path. Unlike other forces, centripetal force is directed towards the center of the circle or curve.

When you swing a ball on a string, the tension in the string provides the inward centripetal force. In the railroad car scenario, the centripetal force acts horizontally to pull the plumb line towards the center of the circular path. This force helps balance the plumb line, preventing it from moving too far outside of the circle.

Mathematically, we calculate the centripetal force using the formula:
  • \( F_c = m \frac{v^2}{r} \)
  • where \( m \) is mass, \( v \) is velocity, and \( r \) is the radius.
Understanding how centripetal force works helps in solving many physics problems involving circular motion.
Gravitational Force
The gravitational force is a force that attracts two bodies towards each other. It is why objects fall towards the Earth and why we stay anchored to the ground. In the context of our physics problem, gravitational force pulls the plumb line straight down.

Gravitational force can be calculated using the equation:
  • \( F_g = mg \)
  • where \( m \) is mass and \( g \) is the acceleration due to gravity (approximately 9.8 m/s虏).
On our plumb line, this force acts vertically. Combined with the horizontal centripetal force, it helps to form the angle of inclination, which is crucial to understanding the problem.
Inclined Plane
An inclined plane, or a slope, is a flat surface tilted at an angle. It simplifies calculations of forces and motion. In our scenario, even though we are dealing with a plumb line rather than a plane, the concept is similar. The inclination of the string is akin to an inclined plane, reflecting the balance of forces acting on the object.

Understanding the inclination helps in visualizing how different forces act. For the plumb line, the angle of inclination tells us how much horizontal force is acting compared to the vertical force. These components ultimately balance to maintain equilibrium during the circular motion of the car.
Physics Problem Solving
Solving physics problems can seem like solving a puzzle. It involves understanding given information and using known equations to find unknown quantities. Here's a simplified approach:

  • Identify what you know and what you need to find.
  • Analyze the forces in the problem; here, they were centripetal and gravitational.
  • Use relevant equations to represent these forces, such as \( F_c = m \frac{v^2}{r} \) for centripetal force.
  • Relate the equations to the angle of inclination using trigonometric relations.
By breaking down the steps and focusing on core concepts, you can approach even complex physics problems with greater confidence.

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Most popular questions from this chapter

A car starts from rest with a constant acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\). Alter \(5 \mathrm{~s}\), a ball is dropped through the window of the ear. The window of the car is at a height of \(1.25 \mathrm{~m}\) from the ground. The angle which the velocity vector makes with the vertical is (a) \(\tan ^{-1}\left(\frac{1}{2}\right)\) (b) \(\tan ^{-1}(2)\) (c) \(\tan ^{-1}\left(\frac{1}{5}\right)\) (d) \(\tan ^{-1}(5)\)

Statement-1: Nomal contact forec offered by a horizontal surface on a block placed on it doesn't form an action-reaction pair with the wcight of the block. Statement-2: Contact force is electromagnetic in nature and weight is gravitational.

Consider a block of mass \(m\) kept on a rough horizontal surfacc. Statement-1 : It is possible to apply a non-vertical force of however large magnitude without disturbing the equilibrium of block. Statement-2: If no force is applied, force of friciton is zero.

Two masses \(m_{1}\) and \(m_{2}\) which are connected with a light string, are placed over a frictionless pulley. This set up is placed over a weighing machine, as shown. Three combination of masses \(m_{1}\) and \(m_{2}\) are used, in first case \(m_{1}=6 \mathrm{~kg}\) and \(m_{2}=2 \mathrm{~kg}\), in second case \(m_{1}=5 \mathrm{~kg}\) and \(m_{2}=3 \mathrm{~kg}\) and in third case \(m_{1}=4 \mathrm{~kg}\) and \(m_{2}=4 \mathrm{~kg}\). Masses are held stationary initially and then released. If the readings of the weighing machinc alter the release in three cases are \(W_{1}, W_{2}\) and \(W_{3}\) respectively then (a) \(W_{1}>W_{2}>W_{3}\) (b) \(W_{1}

Two blocks having masses \(m_{1}\) and \(m_{2}\) are connected by a thread and are placed on an inclined plane with thread initially in a state of no Lension. Thread will not develop any tension if (a) \(m_{1}>m_{2}\) (b) Plane is smooth (c) No friction acts on \(\mathrm{m} 1\) while it acts on \(\mathrm{m}_{2}\) (d) No friction acts on \(m_{2}\) while it acts on \(m_{1}\)
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