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Chapter 6: Problem 60

A pendulum of mass \(m\) hangs from a support fixed to a trolley. The direction of the string when the trolley rolls up a plane of inclination \(\alpha\) with acceleration \(a_{0}\) is (1) \(\theta=\tan ^{-1} \alpha\) (2) \(\theta=\tan ^{-1}\left(\frac{a_{0}}{g}\right)\) (3) \(\theta=\tan ^{-1}\left(\frac{g}{a_{0}}\right)\) (4) \(\theta=\tan ^{-1}\left(\frac{a_{0}+g \sin \alpha}{g \cos \alpha}\right)\)

Short Answer

Expert verified
The correct answer is (4): \(\theta=\tan ^{-1}\left(\frac{a_{0}+g \sin \alpha}{g \cos \alpha}\right)\).

Step by step solution

01

Analyze Forces on the Pendulum

Consider the forces acting on the pendulum. Since the trolley accelerates up a plane with inclination \(\alpha\) and acceleration \(a_0\), inertial forces act on the pendulum. These forces have two components due to gravity and trolley's acceleration.
02

Set Up the Coordinate System

Orient the coordinate system such that the x-axis is along the incline and the y-axis is perpendicular to the incline. The gravitational force can be broken into components: \(mg\cos\alpha\) perpendicular to the incline and \(mg\sin\alpha\) parallel to the incline.
03

Analyze the Inertial Frame

From an inertial frame, consider how the pendulum moves. Due to the trolley鈥檚 acceleration, an apparent force \(ma_0\) acts opposite the direction of the trolley's motion along the incline. This apparent force is horizontal in the inclined frame.
04

Balance the Forces

In equilibrium, the net force must be zero. Let's balance forces along the x and y directions in the non-inertial frame. The horizontal component (co-linear with incline) balances \(ma_0 + mg\sin\alpha\), and the vertical component balances \(mg\cos\alpha\).
05

Calculate Angle \(\theta\)

The angle \(\theta\) of the pendulum with the vertical can be found by the tangent of opposite to adjacent forces: \[ \tan \theta = \frac{ma_0 + mg\sin\alpha}{mg\cos\alpha} \]
06

Solve for \(\theta\)

Simplify the tangent equation: \[ \theta = \tan^{-1}\left(\frac{a_0 + g\sin\alpha}{g\cos\alpha}\right) \]. Comparing with the given options, this matches option (4).

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Key Concepts

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

Inclined Plane
Understanding an inclined plane is pivotal in physics, especially when dealing with motion on a slope. An inclined plane is a flat supporting surface tilted at an angle, allowing objects to move along it either under the influence of gravity or an external force.
The angle of inclination, denoted by \(\alpha\), is critical as it affects the force components acting on any object on the plane. It's crucial to set up a coordinate system aligned with the incline to analyze the forces effectively.
When a trolley moves up an inclined plane with acceleration, analyzing its path requires breaking down gravitational forces and any external acceleration along the inclined and perpendicular directions. This coordinate alignment simplifies force analysis and helps in understanding the motion dynamics of objects like a pendulum attached to a trolley on an incline.
Inertial Forces
In the context of a pendulum in motion on a moving trolley, inertial forces play a significant role. When the trolley accelerates up the inclined plane, inertial forces appear to act opposite to the direction of motion. These are not real physical forces but rather effects of inertia, which is the resistance of any object to a change in its motion.
In this scenario, the pendulum experiences an inertial force equivalent to \(ma_0\), where \(a_0\) is the acceleration of the trolley. This forces the pendulum to shift in a direction that appears opposite to the acceleration. Hence, from a frame of reference on the trolley, the pendulum is affected by this non-inertial force, which is crucial in determining its angle \(\theta\) relative to vertical.
Gravitational Components
Breaking down gravitational components is crucial for understanding forces on an inclined plane. Gravity acts vertically downward, but when you incline the plane, this force must be split into two components relative to the plane:
  • \(mg \cos \alpha\): This component is perpendicular to the inclined plane. It is responsible for the normal force that the plane exerts on the object (perpendicular push back against the object).

  • \(mg \sin \alpha\): This component acts parallel to the inclined plane. It tends to pull the object down along the slope.
By analyzing these components, it becomes easier to understand how objects move on inclined planes. In our context, they help determine the pendulum's equilibrium position by balancing the forces along the slope.
Equilibrium of Forces
The equilibrium of forces involves balancing all acting forces so that an object remains at a constant motion or at rest. In our pendulum exercise, achieving equilibrium means all forces acting in both horizontal and vertical directions must be balanced.
To solve for the pendulum's angle \(\theta\), consider the horizontal and vertical force balances:
  • Horizontal Balance: The force \((ma_0 + mg\sin\alpha)\) opposes the motion due to inertia and gravity along the incline.

  • Vertical Balance: The force \(mg\cos\alpha\) acts perpendicular to the incline, balancing the pendulum's weight component.
In equilibrium, the net force in both directions is zero, allowing us to compute the angle of the pendulum using the tangent function: \(\tan \theta = \frac{ma_0 + mg\sin\alpha}{mg\cos\alpha}\). Solving this provides the accurate configuration of the forces at the observed angle.

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Most popular questions from this chapter

Inside a horizontally moving box, an experimenter finds that when an object is placed on a smooth horizontal table and is released, it moves with an acceleration of \(10 \mathrm{~ms}^{-2}\). In this box, if \(1-\mathrm{kg}\) body is suspended with a light string, the tension in the string in equilibrium position. (w.r.t. experimenter) will be (take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\) ) (1) \(10 \mathrm{~N}\) (2) \(10 \sqrt{2} \mathrm{~N}\) (3) \(20 \mathrm{~N}\) (4) Zero

Figure represents a painter in a crate which hangs alongside a building. When the painter of mass \(100 \mathrm{~kg}\) pulls the rope, the force exerted by him on the floor of the crate is \(450 \mathrm{~N}\). If the crate weighs \(25 \mathrm{~kg}\), find the acceleration (in \(\mathrm{ms}^{-2}\) ) of the painter.

Column I describes the motion of the object and one or more of the entries of Column II may be the cause of motions described in Column I. Match the entries of Column I with the entries of Column II. $$ \begin{array}{|c|c|c|} {\text { Column I }} & {\text { Column II }} \\ \hline \text { i. } \begin{array}{l} \text { An object is moving } \\ \text { towards east. } \end{array} & \begin{array}{c} \text { a. } \\ \text { The net force acting } \\ \text { on the object must } \\ \text { be towards east. } \end{array} \\ \hline \text { ii. } \begin{array}{l} \text { An object is moving } \\ \text { towards east with } \\ \text { constant acceleration. } \end{array} & \text { b. } \begin{array}{l} \text { At least one force } \\ \text { must act towards } \\ \text { east. } \end{array} \\ \hline \text { iii. } \begin{array}{l} \text { An object is moving } \\ \text { towards east with } \\ \text { varying acceleration. } \end{array} & \text { c. } \begin{array}{l} \text { No force may act } \\ \text { towards east. } \end{array} \\ \hline \text { iv. } \begin{array}{l} \text { An object is moving } \\ \text { towards east with } \\ \text { constant velocity. } \end{array} & \text { d. } \begin{array}{l} \text { No force may act } \\ \text { on the object. } \end{array} \\ \hline \end{array} $$

Two identical particles \(A\) and \(B\), each of mass \(m\), are interconnected by a spring of stiffness \(k\). If particle \(b\) experiences a force \(f\) and the elongation of the spring is \(x\) the acceleration of particle \(B\) relative to particle \(A\) is equal to (1) \(\frac{F}{2 m}\) (2) \(\frac{F-k x}{m}\) (3) \(\frac{F-2 k x}{m}\) (4) \(\frac{k x}{m}\)

In order to raise a mass of \(100 \mathrm{~kg}\), a man of mass \(60 \mathrm{~kg}\) fastens a rope to it and passes the rope over a smooth pulley. He climbs the rope with acceleration \(5 g / 4\) relative to the rope. The tension in the rope is (take \(g=10 \mathrm{~ms}^{-2}\) ) (1) \(\frac{4875}{8} \mathrm{~N}\) (2) \(\frac{4875}{2} \mathrm{~N}\) (3) \(\frac{4875}{4} \mathrm{~N}\) (4) \(\frac{4875}{6} \mathrm{~N}\)
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