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Chapter 3: Problem 40

\(\Lambda\) particle moves in a circle with increasing speed \(v=k t\) where \(k\) is a positive constant. Find the total acceleration of the particle when it revolves \(n\) tines. \(\Lambda\) ssume \(r=\) radius of the circle

Short Answer

Expert verified
The total acceleration after revolving n times is \(\sqrt{k^2 + \frac{4n^2\pi^2 k^2}{r^2}}\).

Step by step solution

01

Determine the Angular Speed

The speed of the particle is given by k , and it increases linearly with time . The angular speed of the particle is found by dividing the linear speed by the radius of the circle: \[\omega = \frac{v}{r} = \frac{kt}{r}.\]
02

Calculate Tangential Acceleration

The tangential acceleration is the rate of change of linear speed. Since the linear speed is \(kt\), its derivative with respect to time gives us the tangential acceleration:\[a_t = \frac{d(kt)}{dt} = k.\]
03

Calculate Centripetal Acceleration

The centripetal acceleration is given by \[a_c = \frac{v^2}{r} = \frac{(kt)^2}{r} = \frac{k^2 t^2}{r}.\]
04

Determine Total Acceleration

The total acceleration of a particle moving in a circle with both radial and tangential components is given by:\[a = \sqrt{a_t^2 + a_c^2}.\]Substituting the expressions of tangential and centripetal acceleration, we get:\[a = \sqrt{k^2 + \frac{k^4 t^4}{r^2}}.\]
05

Substitute the Time for Complete Revolutions

To find the time it takes to complete n revolutions, set the number of revolutions equal to the angular displacement:\[n \cdot 2\pi = \int_{0}^{t} \omega dt = \int_{0}^{t} \frac{kt}{r} dt = \frac{k}{r} \cdot \frac{t^2}{2}.\]Solving for \(t\) gives:\[t = \sqrt{\frac{2n\pi r}{k}}.\]
06

Determine Total Acceleration After n Revolutions

Substitute the time from Step 5 into the expression for total acceleration:\[a = \sqrt{k^2 + \frac{k^4 \left(\sqrt{\frac{2n\pi r}{k}}\right)^4}{r^2}}\]Simplify this expression:\[= \sqrt{k^2 + \frac{k^4 \left(\frac{2n\pi r}{k}\right)^2}{r^2}}\]\[= \sqrt{k^2 + \frac{k^4 \cdot \frac{4n^2\pi^2 r^2}{k^2}}{r^2}}\]\[= \sqrt{k^2 + \frac{4n^2\pi^2 k^2}{r^2}}.\]

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Key Concepts

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

Total Acceleration
In a circular motion, a particle's acceleration is due to two primary components: tangential and centripetal acceleration. To find the total acceleration, we combine both using the Pythagorean theorem since they are perpendicular to each other. For our particle, its total acceleration can be represented by the equation:
  • \[ a = \sqrt{a_t^2 + a_c^2} \]
  • Where \( a_t \) is the tangential acceleration and \( a_c \) is the centripetal acceleration.
This gives us a comprehensive figure that captures both the change in speed and the direction change of the particle as it moves around the circle.
To apply this, substitute the values you have for both types of acceleration into this formula to find out how much the movement is affecting the particle overall as it continues its path in the circular motion.
Angular Speed
Angular speed is a critical element in understanding circular motion. It tells us how fast our particle is spinning around the circle. Mathematically, the angular speed \( \omega \) is obtained by dividing the linear speed \( v \) of a particle by the radius \( r \) of the path:
  • \[ \omega = \frac{v}{r} \]
  • For our problem, this becomes \( \omega = \frac{kt}{r} \).
Angular speed can vary with time if the linear speed of the particle changes, and here it increases linearly with time. Understanding and calculating angular speed is essential because it is directly tied to many other dynamic quantities like angular displacement and tangential velocity.
Centripetal Acceleration
Centripetal acceleration is what keeps a particle moving in a circular path. It acts towards the center of the circle and is responsible for changing the direction of the velocity. Its magnitude can be determined using the formula:
  • \[ a_c = \frac{v^2}{r} \]
  • In the scenario of the exercise, this becomes \( a_c = \frac{k^2 t^2}{r} \).
This expression tells us how the acceleration changes with speed and time. Importantly, centering this acceleration means it does not affect the tangential speed but ensures the path remains circular.
Understanding centripetal acceleration is key to any problem involving circular motion, especially when motion is not uniform.
Tangential Acceleration
Tangential acceleration refers to how quickly the speed of a particle moving along the circular path changes. It's akin to the 'ramping up' of the speed as the particle moves around the circle, enhancing or reducing the speed at any point:
  • In this exercise, we use \( a_t = \frac{d(kt)}{dt} \) which simplifies to \( a_t = k \).
Purely, it signifies how much the particle's speed is changing per unit of time, and unlike centripetal acceleration, it acts along the tangent to the circle.
This clear distinction helps when solving problems since you can consider the two accelerations independently before combining them into the total acceleration for a fuller picture.

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Most popular questions from this chapter

\Lambda particle moving in \(x-y\) plane is at origin at time \(t=0 .\) Velocity of the particle is \((2 \vec{i}-4 \hat{j}) \mathrm{m} / \mathrm{s}\) and acceleration \(i s(4 \hat{i} \mid \hat{j}) \mathrm{m} / \mathrm{s}^{2}\). Find at \(t=2 \mathrm{~s}\) (a) velocity of particle and (b) coordinates of particle.

A particle is projecled from a horizontal plane with speed \(u\) at some ungle. At highest point its velocity is found to be \(u / 2\). The maximum height of the projcctile will be (a) \(\frac{u^{2}}{4 g}\) (b) \(\frac{3 u^{2}}{4 g}\) (c) \(\frac{3 u^{2}}{8 g}\) (d) \(\frac{u^{2}}{8 g}\)

Mass \(A\) is released from the top of a frictionless inclined plane \(18 \mathrm{~m}\) long and reaches the bottom 3 s later: \(\Lambda\) t the instant when \(A\) is released, a second mass \(B\) is projected upwards along the plane from the bottom with a certain initial velocity. Mass \(B\) travels a distance up the plane, stops and returns to the bottom so that it arrives simultaneously with \(A\). The two masses do not collide. Initial velocity of \(A\) is (a) \(4 \mathrm{~ms}^{1}\) (b) \(5 \mathrm{~ms}^{1}\) (c) \(6 \mathrm{~ms}\) (d) \(7 \mathrm{~ms}^{1}\)

In uniform circular motion a body moves with constant speed \(v\) on a circular path of constant radius \(r\). In this motion Column-I (a) The acceleration of body is (b) The kinetic energy (c) The angular displacement of body at any instant is directed along (d) The velocity of body is always directed along Column-II (p) tangent to path at evety point (q) alng axis of rotation (r) constant in magnitude but changing in direction (s) constant always

\(\Lambda\) particle is projected from the bottom of an inclined plane of inclination \(30^{\circ}\) with velocity of \(40 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) with horizontal. Find the speed of the particle when its velocity vector is parallel to the plane. 'l'ake \(g=10 \mathrm{~m} / \mathrm{s}^{2}\)
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