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Chapter 10: Problem 17

A particle of mass \(m\) is projected with a speed \(u\) at an angle \(\theta\) with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.

Short Answer

Expert verified
The torque is \( \frac{1}{2} mu^2 \sin(2\theta) \).

Step by step solution

01

Understand the Problem

We need to find the torque exerted by the gravitational force on a particle about the point of projection when it is at its highest point in projectile motion. Torque is the rotational analog of force.
02

Recall the Formula for Torque

Torque \( \tau \) is given by the product of the force, the distance from the pivot point, and the sine of the angle between the force vector and the lever arm. The formula is \( \tau = r F \sin(\varphi) \), where \( r \) is the perpendicular distance from the line of action of the force to the pivot, \( F \) is the force, and \( \varphi \) is the angle between \( r \) and \( F \).
03

Calculate the Distance at the Highest Point

First, calculate the horizontal distance \( r \) to the point of interest when the particle is at its highest point. The horizontal component of velocity is \( u\cos\theta \), and the particle reaches its highest point when the vertical component of its velocity is zero. The time \( t \) to reach the highest point can be found using the equation for vertical velocity: \( 0 = u \sin\theta - gt \). Solve for \( t \):\[ t = \frac{u \sin\theta}{g} \]The horizontal distance \( r = (u\cos\theta) \cdot t = u\cos\theta \cdot \frac{u \sin\theta}{g} \).
04

Calculate the Force

The force acting on the particle is its weight \( W = mg \), and at the highest point, the weight acts vertically downwards.
05

Determine the Angle and Torque

At the highest point of the projectile, the angle \( \varphi \) between the gravitational force and the horizontal displacement is \( 90^\circ \). Now, substitute into the torque formula:\[ \tau = rW\sin(90^\circ) = mgr \]Substitute \( r = \frac{u^2\sin\theta\cos\theta}{g} \) from Step 3:\[ \tau = mg \left( \frac{u^2 \sin\theta \cos\theta}{g} \right) = mu^2 \sin\theta \cos\theta \]Recognize \( \sin\theta \cos\theta = \frac{1}{2} \sin(2\theta) \):\[ \tau = \frac{1}{2} mu^2 \sin(2\theta) \]
06

Conclude the Solution

The torque of the weight of the particle about the point of projection, when the particle is at the highest point, is given by: \[ \tau = \frac{1}{2} mu^2 \sin(2\theta) \]

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Key Concepts

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

Torque
Torque is a concept that describes how forces affect rotational motion. Imagine opening a door: the force you apply, and where you push, matters. This is where torque comes in.
  • Torque measures the effectiveness of a force in causing an object to rotate around a pivot point.
  • The direction and size of the force, along with the distance from the pivot, all play roles.
The formula for torque (\( \tau \)) is: \[ \tau = r F \sin(\varphi) \]where \( r \)is the distance from the line of force to the pivot, \( F \) is the force, and \( \varphi \)is the angle between the force and the lever. Torque is greater when you apply force farther from the pivot or at the right angle. Think of how a wrench works better at a right angle to the bolt!
Gravitational Force
Gravitational force is something we feel every day. It's the force that pulls objects towards the Earth, or any other large body like a planet.
  • This force is proportional to the mass of the objects and inversely proportional to the square of the distance between their centers.
  • Near the Earth's surface, it acts downwards with a force equal to the weight of an object, calculated as \( W = mg \), where \( m \)is mass, and \( g \)is the acceleration due to gravity (approximately 9.8 \, m/s^2).
In projectile motion, the gravitational force causes the object to follow a parabolic path by constantly accelerating it downwards.
Highest Point in Trajectory
The highest point in a projectile's trajectory is crucial to understanding its motion.
  • At this point, the vertical component of the projectile's velocity is zero.
  • The projectile only has horizontal motion, as gravity has zero influence on horizontal motion directly.
At this highest point, the projectile is farthest from the ground, and its motion is temporarily no longer upwards. Calculating this point involves finding when the velocity component perpendicular to the ground becomes zero. In equations, this results in interesting dynamics, such as calculating the time to reach this point using the equation \( 0 = u \sin\theta - gt \).Understanding this concept helps in calculating things like the range or the total time of flight.
Rotational Motion
Rotational motion is the movement of objects around an axis. It's different from linear motion in many aspects and is a key concept in physics.
  • In rotational motion, angular velocity and angular displacement describe how fast and how far an object rotates.
  • Torque, as mentioned earlier, is the force that causes this motion, just as a push or pull causes linear motion.
When examining projectiles, like in this exercise, we study how different forces, particularly gravity, can cause a change in the motion trajectory without causing rotation, but forces that act off-center can certainly induce such motion.

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Most popular questions from this chapter

A kid of mass \(M\) stands at the edge of a platform of radius \(R\) which can be freely rotated about its axis. The moment of inertia of the platform is \(I\). The system is at rest when a friend throws a ball of mass \(m\) and the kid catches it. If the velocity of the ball is \(v\) horizontally along the tangent to the edge of the platform when it was caught by the kid, find the angular speed of the platform after the event.

A block hangs from a string wrapped on a disc of radius \(20 \mathrm{~cm}\) free to rotate about its axis which is fixed in a horizontal position. If the angular speed of the disc is \(10 \mathrm{rad} / \mathrm{s}\) at some instant, with what speed is the block going down at that instant?

A uniform rod of mass \(300 \mathrm{~g}\) and length \(50 \mathrm{~cm}\) rotates at a uniform angular speed of \(2 \mathrm{rad} / \mathrm{s}\) about an axis perpendicular to the rod through an end. Calculate (a) the angular momentum of the rod about the axis of rotation, (b) the speed of the centre of the rod and (c) its kinetic energy.

A cylinder rotating at an angular speed of \(50 \mathrm{rev} / \mathrm{s}\) is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed?

Find the moment of inertia of a uniform square plate of mass \(m\) and edge \(a\) about one of its diagonals.
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