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Chapter 9: Problem 9

The angular velocity of a particle moving in a circle radius \(50 \mathrm{~cm}\) is increased in 5 min from 100 revolutions \(p\) minute to 400 revolutions per minute. Find the tangent acceleration of the particle. (1) \(60 \mathrm{~ms}^{-2}\) (3) \(\pi / 15 \mathrm{~ms}^{-2}\) (2) \(\pi / 30 \mathrm{~ms}^{-2}\) (4) \(\pi 60 \mathrm{~ms}^{-2}\)

Short Answer

Expert verified
The tangent acceleration of the particle is \(\frac{\pi}{60} \mathrm{~ms}^{-2}\).

Step by step solution

01

Understand the Problem

We need to find the tangent acceleration of a particle moving in a circular path. The angular velocity of the particle is initially 100 revolutions per minute and is increased to 400 revolutions per minute over 5 minutes.
02

Convert Angular Velocity to Radians per Second

First, we'll convert the initial and final angular velocities from revolutions per minute to radians per second. Remember that 1 revolution is equal to \(2\pi\) radians.- Initial angular velocity: \(\omega_1 = 100 \text{ revolutions/minute} \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} = \frac{100 \times 2\pi}{60} \text{ rad/s} = \frac{10\pi}{3} \text{ rad/s}\)- Final angular velocity: \(\omega_2 = 400 \text{ revolutions/minute} \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} = \frac{400 \times 2\pi}{60} \text{ rad/s} = \frac{40\pi}{3} \text{ rad/s}\)
03

Calculate Angular Acceleration

Using the formula for angular acceleration, \(\alpha\), which is \(\alpha = \frac{\Delta \omega}{\Delta t}\), calculate the angular acceleration:\[\alpha = \frac{\left(\frac{40\pi}{3} - \frac{10\pi}{3}\right) \text{ rad/s}}{5 \times 60 \text{ s}} = \frac{30\pi}{3 \times 300} \text{ rad/s}^2 = \frac{\pi}{30} \text{ rad/s}^2\]
04

Find Tangent Acceleration

The tangent acceleration, \(a_t\), of the particle is given by: \(a_t = r \cdot \alpha\), where \(r = 50\text{ cm} = 0.5\text{ m}\).\[a_t = 0.5 \cdot \frac{\pi}{30} = \frac{\pi}{60} \text{ m/s}^2\]
05

Compare with Given Options

Check which option from the problem matches the calculated tangent acceleration value. The correct option is (2) \(\pi / 30 \mathrm{~ms}^{-2}\).

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Key Concepts

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

Angular Velocity
Angular velocity is a fundamental concept in circular motion that describes how quickly an object rotates around a circle. It's similar to linear velocity but applies to rotational or circular paths rather than straight lines. Angular velocity is denoted by the symbol \( \omega \). It is typically measured in radians per second (rad/s), though it can also be expressed in revolutions per minute (RPM).

Key things to consider about angular velocity include:
  • Directional Aspect: The direction of angular velocity is determined by the right-hand rule. If the fingers of your right hand follow the direction of rotation, your thumb points in the direction of angular velocity.
  • Calculation: To convert from revolutions per minute to radians per second, use the relation: \( 1 \text{ revolution} = 2\pi \text{ radians} \). Multiply by \( \frac{1\text{ minute}}{60\text{ seconds}} \) to switch the time unit to seconds.
  • Significance: Understanding angular velocity is crucial because it helps determine other dynamic properties such as angular acceleration and tangential acceleration in circular motion.
Tangent Acceleration
Tangent acceleration, also known as tangential acceleration, is the rate of change of tangential velocity as an object moves along a circular path. It is responsible for the change in the speed (not the direction) of the object moving along the circumference.

Characteristics of tangent acceleration:
  • Dependency on Radius: Tangent acceleration depends on both the angular acceleration and the radius of the circular path. It is calculated using the formula \( a_t = r \cdot \alpha \), where \( r \) is the radius and \( \alpha \) is the angular acceleration.
  • Measurement: The unit of tangent acceleration is meters per second squared (m/s虏), indicating how fast the tangential speed is increasing or decreasing.
  • Relation to Angular Quantities: Changes in angular velocity affect tangential acceleration directly. If there is angular acceleration present, there will be a corresponding tangent acceleration influencing the linear speed.
Angular Acceleration
Angular acceleration refers to the rate of change of angular velocity over time. It tells us how quickly an object is speeding up or slowing down its rotation. Angular acceleration is a key concept when analyzing situations where an object's rotational speed is not constant.

Key insights into angular acceleration:
  • Units and Symbols: Angular acceleration is denoted by \( \alpha \) and is measured in radians per second squared (rad/s虏).
  • Link to Angular Velocity: It is calculated using the formula \( \alpha = \frac{\Delta \omega}{\Delta t} \), where \( \Delta \omega \) is the change in angular velocity and \( \Delta t \) is the change in time.
  • Influence on Tangential Motion: As angular acceleration increases, so does the tangential acceleration, affecting the object's linear speed along the circular path.
  • Practical Importance: Angular acceleration is essential for understanding the dynamics of rotating wheels, gears, or celestial bodies, where changes in rotational motion need to be quantified.

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Most popular questions from this chapter

A stone is thrown from point \(O\) on ground with velocity \(\vec{v}=5 \hat{i}+10 \hat{j} \mathrm{~m} / \mathrm{s}\), where \(x\) is horizontal and \(y\) is vertical. Its. angular velocity about \(O\) when the stone is at maximum height of its trajectory, is (i) \(\frac{1}{2} \mathrm{rad} / \mathrm{s}\) (2) \(\frac{1}{\sqrt{2}} \mathrm{rad} / \mathrm{s}\) (3) \(1 \mathrm{rad} / \mathrm{s}\) (4) \(\sqrt{2} \mathrm{rad} / \mathrm{s}\)

A particle of mass \(m\) moves along a circular path of radius \(r\) with a centripetal acceleration \(a_{n}\) changing with time \(t\) as \(a_{n}=k t^{2}\), where \(k\) is a positive constant. The average power developed by all the forces acting on the particle during the first \(t_{0}\) seconds is (1) \(m k r t_{0}\) (2) \(\frac{m k r t_{0}^{2}}{2}\) (3) \(\frac{m k r t_{0}}{2}\) (4) \(\frac{m k r t_{0}}{4}\)

A heavy particle is tied to the end \(A\) of a string of length \(1.6 \mathrm{~m}\). Its other end \(O\) is fixed. It revolves as a conical pendulum with the string making \(60^{\circ}\) with the vertical. Then (1) its period of revolution is \(4 \pi / 7 \mathrm{~s}\). (2) the tension in the string is doubled the weight of the particle (3) the velocity of the particle \(=2.8 \sqrt{3} \mathrm{~m} / \mathrm{s}\) (4) the centripetal accelcration of the particle is \(9.8 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\).

A particle tied at one end of an inextensible string of length \(R\), whose another end is fixed. Particle is performing circular motion in vertical plane. Ratio of maximum to minimum speed is \(\lambda\). Choose the correct option(s). (1) Speed at the top most point is \(\sqrt{\left(\frac{4 \lambda^{2}}{\lambda^{2}-1}\right) g R}\) (2) Speed at the top most point is \(\sqrt{\left(\frac{4}{\lambda^{2}-1}\right) g R}\) (3) Ratio of maximum to minimum tension is \(\left(\frac{5 \lambda^{2}-1}{5-\lambda^{2}}\right)\) (4) Ratio of tension in string to weight when string is horizontal is \(2\left(\frac{\lambda^{2}-1}{\lambda^{2}+1}\right)\)

A small sphere is given vertical velocity of magnitude \(v_{0}=5 \mathrm{~ms}^{-1}\) and it swings in a vertical plane about the end of a massless string. The angle \(\theta\) with the vertical at which string will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere, is (1) \(\cos ^{-1}\left(\frac{2}{3}\right)\) (2) \(\cos ^{-1}\left(\frac{1}{4}\right)\) (3) \(60^{\circ}\) (4) \(30^{\circ}\)
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