91Ó°ÊÓ

Find the acceleration of a small body \(A\) which slides without initial velocity down a helical groove with a pitch \(h\) and radius \(R\) at the end of the \(n^{\text {th }}\) turn Neglect friction. Solution In \(n\) turns the particle will fall through a vertical height \(n h\). So, by the conservation of energy $$ m g(n h)-\frac{1}{2} m v^{2} \Rightarrow v-\sqrt{2 n g h} $$ The particle will have two accelerations at any instant: the tangential acceleration \(a_{t}=g \sin \alpha\), where \(\alpha=\) inclination of the groove with the horizontal and the normal acceleration $$ a_{n}-\frac{(v \cos \alpha)^{2}}{R} $$To find \(\alpha\), we mentally flatten the surlace of the cylinder with the helical groove into a plane surface. The groove will become the hypotenuse of a right-angled triangle of height \(n h\) and base \(2 \pi n R\) and so $$ \begin{aligned} & \tan \alpha-\frac{n h}{2 \pi n R}-\frac{h}{2 \pi R} \\ \therefore & a-\sqrt{a_{t}^{2}+a_{n}^{2}}-\sqrt{g^{2} \sin ^{2} \alpha+\frac{v^{4} \cos ^{4} \alpha}{R^{2}}} \\ \Rightarrow & a-\sqrt{g^{2} \frac{h^{2}}{h^{2}+4 \pi^{2} R^{2}}+\frac{4 n^{2} g^{2} h^{2}}{R^{2}} \frac{16 \pi^{4} R^{4}}{\left(h^{2}+4 \pi^{2} R^{2}\right)^{2}}} \\ \Rightarrow & a-\frac{g h \sqrt{h^{2} \mid 4 \pi^{2} R^{2}+64 \pi^{4} n^{2} R^{2}}}{h^{2}+4 \pi^{2} R^{2}} \end{aligned} $$

Step by step solution

01

Determine Vertical Height After n Turns

In the helix, after completing n turns, the particle has descended vertically by a height of \(nh\). This height is crucial for determining the dynamics of the particle as it relates to its potential energy change.

02

Apply Conservation of Energy

The initial potential energy \(mgh\) of the particle is converted into kinetic energy as it descends. The conservation of energy equation becomes:\[mgn h = \frac{1}{2} m v^2\]Solving for \(v\) gives:\[v = \sqrt{2ngh}\]

03

Define Tangential and Normal Accelerations

At any point, the particle experiences a tangential acceleration \(a_t = g \sin\alpha\) and a normal acceleration due to the curvature of the path \(a_n = \frac{(v\cos\alpha)^2}{R}\).

04

Calculate Angle of Inclination \(\alpha\)

Unwind the helical path onto a flat surface, which forms a right triangle. The height is \(nh\), and the base is \(2\pi nR\), giving:\[\tan\alpha = \frac{nh}{2\pi nR} = \frac{h}{2\pi R}\]

05

Replace in Tangential Acceleration Formula

Substituting \(\alpha\) into \(a_t\) gives:\[a_t = g \sin \alpha = g \frac{h}{\sqrt{h^2 + (2\pi R)^2}}\]

06

Replace in Normal Acceleration Formula

Insert \(v\) and \(\alpha\) into the normal acceleration formula:\[a_n = \frac{(v\cos\alpha)^2}{R} = \frac{(\sqrt{2ngh}\cos\alpha)^2}{R}\]\[a_n = \frac{2ngh \frac{4\pi^2R^2}{h^2 + 4\pi^2R^2}}{R}\]

07

Compute the Total Acceleration

The resultant acceleration \(a\) is the combination of the tangential and normal components:\[a = \sqrt{a_t^2 + a_n^2} = \sqrt{g^2 \sin^2 \alpha + \frac{(v\cos\alpha)^4}{R^2}}\]Insert the expressions for \(a_t\) and \(a_n\):\[a = \frac{gh \sqrt{h^2 + 4\pi^2R^2 + 64\pi^4n^2R^2}}{h^2 + 4\pi^2R^2}\]

08

Final Step: Simplify the Expression for Acceleration

Finally, the simplified form of the acceleration taking into account all components is:\[a = \frac{gh \sqrt{h^2 + 4\pi^2R^2 + 64\pi^4n^2R^2}}{h^2 + 4\pi^2R^2}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Conservation of Energy

Understanding the conservation of energy is crucial in problems involving motion along a helical path without friction. When a small body slides down a helical groove, its potential energy decreases while its kinetic energy increases. At the beginning of its journey, the particle has maximum potential energy, which is converted into kinetic energy as the particle slides down. This concept is expressed through the equation:

• Initial Potential Energy: \( mgnh \)
• Kinetic Energy at Any Point: \( \frac{1}{2}mv^2 \)

By conserving energy, we can infer that the particle's velocity at the nth turn of the helix is given by \( v = \sqrt{2ngh} \), derived from setting the initial potential energy equal to the kinetic energy the particle gains at that point. This transition from potential to kinetic energy describes how energy conservation underlines the change from being at rest to reaching a certain velocity.

Tangential Acceleration

The force that causes the particle to slide down the helix is governed by tangential acceleration. This acceleration is parallel to the incline's surface and depends on the incline's angle. To calculate tangential acceleration, use the formula:

• \( a_t = g \sin \alpha \)

where \( \alpha \) is the angle of inclination derived from the geometry of the problem. Using trigonometry, the incline angle is determined by unwinding the helix into a right triangle. Here, the angle \( \alpha \) is calculated by the ratio of the height of a helix turn to its base, given by \( \tan \alpha = \frac{h}{2\pi R} \). This leads us to use \( \sin \alpha = \frac{h}{\sqrt{h^2 + (2\pi R)^2}} \) in finding \( a_t \). The geometry and trigonometry here help pinpoint the effect of gravity on the body as it moves.

Normal Acceleration

Normal acceleration acts perpendicularly to the path of motion and is caused by the helical shape of the groove, keeping the particle on its curving path. Unlike tangential acceleration, it doesn't increase or decrease the object's speed directly but maintains its circular path by providing a centripetal force. The formula for normal acceleration is:

• \( a_n = \frac{(v \cos \alpha)^2}{R} \)

By substituting in the velocity derived earlier, \( v = \sqrt{2ngh} \), and considering the path's geometry for \( \cos \alpha \), this becomes:

• \( a_n = \frac{2ngh \cdot \frac{4\pi^2R^2}{h^2 + 4\pi^2R^2}}{R} \)

Normal acceleration ensures that as the particle speeds up or slows down along its path around the helix, it stays on the helical track and does not deviate.

Inclined Plane Analysis

Inclined plane analysis is key to understand the dynamics of motion in a helical groove. When the cylinder with the helical groove is mentally flattened, it forms a right-angled triangle, allowing for applications of trigonometry to determine critical angles and accelerations. In this configuration:

• Height of the Triangle: \( nh \)
• Base of the Triangle: \( 2\pi nR \)

These dimensions help determine the incline angle \( \alpha \) crucial for both tangential and normal accelerations. Calculating \( \tan \alpha = \frac{h}{2\pi R} \) and finding \( \sin \alpha \) and \( \cos \alpha \) lead to understanding how the forces act on the sliding body. Exploiting the geometry of the inclined plane reduces a three-dimensional problem into simpler calculations, enabling efficient analysis of the particle's motion.