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Chapter 3: Problem 105

A wheel of radius \(0.20 \mathrm{~m}\) is accelerated from rest with an angular acceleration of \(1 \mathrm{rad} / \mathrm{s}^{2}\). After a rotation of \(90^{\circ}\) the radial acceleration of a particle on its rim will be (a) \(\pi \mathrm{m} / \mathrm{s}^{2}\) (b) \(0.5 \pi \mathrm{m} / \mathrm{s}^{2}\) (c) \(2.0 \pi \mathrm{m} / \mathrm{s}^{2}\) (d) \(0.2 \pi \mathrm{m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The radial acceleration of a particle on its rim after a rotation of \(90^\circ\) is \(0.2\pi \mathrm{m/s^2}\) (d).

Step by step solution

01

Convert Degrees to Radians

First conversion from degrees to radians has to be made as the wheel rotates an angle of \(90^\circ\). In radians, this is \(\frac{\pi}{2}\) rad, because \(360^\circ\) is equivalent to \(2\pi\) rad.
02

Calculate the Angular Velocity

Angular velocity (\(\omega\)) after the rotation can be calculated with the equation \(\omega^2 = 2 \alpha \theta\), where \(\alpha\) is the angular acceleration and \(\theta\) is the angle in radians. Plugging in the values \(\alpha = 1 \mathrm{rad/s^2}\) and \(\theta = \frac{\pi}{2}\) rad, we get \(\omega = \sqrt{2*1*\frac{\pi}{2}} = \sqrt{\pi}\) rad/s.
03

Calculate Radial Acceleration

Now, we can calculate the radial acceleration (\(a_r\)) using the equation \(a_r = \omega^2 r\), where \(r\) is the radius. Plugging in the values \(\omega = \sqrt{\pi}\) rad/s and \(r = 0.20\) m, we get \(a_r = (\sqrt{\pi})^2 * 0.20 = \pi * 0.20 = 0.2\pi\mathrm{m/s^2}\).

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Key Concepts

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

Radial Acceleration
Radial acceleration, often termed centripetal acceleration, is the acceleration experienced by an object moving in a circular path. It points towards the center of the circle and is responsible for changing the direction of the velocity of the object rather than its magnitude.

To calculate radial acceleration (\( a_r \)), we use the formula:
  • \( a_r = \omega^2 r \)
where \( \omega \) is the angular velocity and \( r \) is the radius of the circular path. This formula shows that radial acceleration depends on the square of the angular velocity and the radius of the circle.

For instance, in the exercise discussed, the wheel after a \( 90^{\circ} \) rotation has a radial acceleration of \( 0.2\pi \mathrm{m/s^2} \), which was determined by using this specific relationship.
Angular Velocity
Angular velocity is the rate at which an object rotates or revolves around a central point. It is symbolized by \( \omega \) and measured in radians per second (rad/s). This parameter tells us how quickly an object covers an angle during its motion.

In the given exercise context, after the wheel rotates \( 90^{\circ} \), we calculate angular velocity using the equation:
  • \( \omega^2 = 2 \alpha \theta \)
This equation stems from the principles of rotational kinematics, where \( \alpha \) is the angular acceleration and \( \theta \) the angular displacement in radians.

Substituting the values, we found \( \omega = \sqrt{\pi} \) rad/s. Understanding this allows us to compute the radial acceleration quite accurately.
Angle Conversion to Radians
Angle conversion to radians is a fundamental concept when dealing with rotational motion. Radians provide a universal standard of measurement in physics because they relate angles directly to distances traveled on a circle.

To convert degrees to radians, we use the relation that \( 360^{\circ} = 2\pi \) rad. Thus, any angle in degrees can be converted to radians using the formula:
  • \( \text{{Radians}} = \text{{Degrees}} \times \frac{\pi}{180} \)
This means that a \( 90^{\circ} \) angle becomes \( \frac{\pi}{2} \) rad.

When solving physics problems, using radians simplifies calculations, such as those involving angular velocity and radial acceleration. Always check the angle units during computations to ensure accuracy.

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Most popular questions from this chapter

A body of mass \(100 g\) is rotating in a circular path of radius \(r\) with constant velocity. The work done in one complete revolution is (a) 10or Joule (b) \((r / 100)\) Joule (c) (100/r) Joule (d) Zero

If the equation for the displacement of a particle moving on a circular path is given by \((\theta)=2 t^{3}+0.5\), where \(\theta\) is in radians and \(t\) in seconds, then the angular velocity of the particle after \(2 \mathrm{sec}\) from its start is [AIIMS 1998] (a) \(8 \mathrm{rad} / \mathrm{sec}\) (b) \(12 \mathrm{rad} / \mathrm{sec}\) (c) \(24 \mathrm{rad} / \mathrm{sec}\) (d) \(36 \mathrm{rad} / \mathrm{sec}\)

A body of mass of \(100 \mathrm{~g}\) is attached to a \(1 \mathrm{~m}\) long string and it is revolving in a vertical circle. When the string makes an angle of \(60^{\circ}\) with the vertical then its speed is \(2 \mathrm{~m} / \mathrm{s}\). The tension in the string at \(\theta=60^{\circ}\) will be (a) \(89 \mathrm{~N}\) (b) \(0.89 \mathrm{~N}\) (c) \(8.9 \mathrm{~N}\) (d) \(0.089 \mathrm{~N}\)

A particle of mass \(m\) is describing a circular path of radius \(r\) with uniform speed. If \(L\) is the angular momentum of the particle about the axis of the circle, the kinetic energy of the particle is given by [CPMT 1995] (a) \(L^{2} / m r^{2}\) (b) \(L^{2} / 2 m r^{2}\) (c) \(2 L^{2} / m r^{2}\) (d) \(m r^{2} L\)

A particle moves with a constant speed \(v\) along a circular path of radius \(r\) and completes the circle in time \(T\). What is the acceleration of the particle \(\quad\) [Orissa JEE 2002] (a) \(m g\) (b) \(\frac{2 \pi v}{T}\) (c) \(\frac{\pi r^{2}}{T}\) (d) \(\frac{\pi v^{2}}{T}\)
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