Solved Problem on Kinematics
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Determine the acceleration vector of a body that slides from the rest by a helicoid channel with pitch
k and radius R at the end of the n-th turns, the friction is neglected.
Problem data:
- Initial speed: v0 = 0,
- Helicoid radius: R;
- Helicoid pitch: k.
We choose a cylindrical coordinate system (Figure 1-A), where \( {\mathbf{e}}_{r} \), \( {\mathbf{e}}_{z} \) and \( {\mathbf{e}}_{\theta } \) are the unit vectors of directions r, z, and θ.
The acceleration vector a of the body points to the center of the helicoid in the downward direction. This vector can be decomposed into three components (Figure 1-B), in the z direction the \( -{\mathbf{a}}_{z} \) component pointing down, in the direction r to component \( -{\mathbf{a}}_{r} \) pointing to the center of the curve, and in the direction θ the component \( {\mathbf{a}}_{\theta } \) tangent to the curve.
Solution
The acceleration vector is given by
\[
\begin{gather}
\mathbf{a}=-a_{r}{\mathbf{e}}_{r}+a_{\theta}{\mathbf{e}}_{\theta}-a_{z}{\mathbf{e}}_{z} \tag{I}
\end{gather}
\]
The magnitude of the component in the r direction represents the centripetal acceleration of the body
given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{a_{r}=\frac{v^{2}}{r}} \tag{II}
\end{gather}
\]
Unfolding a helicoid turn, pitch k represents one side of a right triangle, the projection of the
helicoid in a plane is a circle with radius R, which unfolded has a length 2πR, represents
another side (Figure 2). Using Pythagorean Theorem the hypotenuse will be
\[
\begin{gather}
h^{2}=k^{2}+(2\pi R)^{2}\\
h=\sqrt{k^{2}+4\pi^{2}R^{2}\;} \tag{III}
\end{gather}
\]
which represents the length of one turn of the helicoid.
From Figure 2, we have the following relationships
\[
\begin{gather}
\cos \alpha =\frac{2\pi R}{\sqrt{k^{2}+4\pi^{2}R^{2}\;}} \tag{IV-a}
\end{gather}
\]
\[
\begin{gather}
\operatorname{sen}\alpha =\frac{k}{\sqrt{k^{2}+4\pi^{2}R^{2}\;}} \tag{IV-b}
\end{gather}
\]
The speed of the body is tangent to the helicoid (Figure 3-A), decomposing the speed in directions θ
and z, we have the components vθ and vz. The angle between
the velocity vector vT, parallel to the inclined plane, and the θ direction
is α, the same angle of inclination of the plane, these angles are alternate interior angles
(Figure 3-B).
The component vθ is tangent to the helicoid, this component is used for the calculation of the centripetal acceleration given by the expression (II)
\[
v_{\theta }=v_{T}\cos \alpha
\]
substituting v = vθ and r = R into expression (II)
\[
\begin{gather}
a_{r}=\frac{v_{\theta }^{2}}{R}\\
a_{r}=\frac{(v_{T}\cos\theta )^{2}}{R}\\
a_{r}=\frac{v_{T}^{2}\cos ^{2}\theta}{R}
\end{gather}
\]
using the cosine value obtained in (IV)
\[
\begin{gather}
a_{r}=\frac{v_{T}^{2}}{R}\left(\frac{2\pi R}{\sqrt{k^{2}+4\pi^{2}R^{2}\;}}\right)^{2}\\
a_{r}=\frac{v_{T}^{2}}{R}\frac{4\pi^{2}R^{2}}{k^{2}+4\pi ^{2}R^{2}}\\
a_{r}=v_{T}^{2}\frac{4\pi^{2}R}{k^{2}+4\pi ^{2}R^{2}} \tag{V}
\end{gather}
\]
The tangential speed vT is found using the
Principle of Conservation of Mechanical Energy. Unfolding all n helicoid laps we will have the height
of the inclined plane equal to n.k. The initial mechanical energy
(\( E_{i} \)),
at the top of the plane, is equal to the final mechanical energy
(\( E_{f} \)),
at the base of the plane (Figure 4).
\[
E_{i}=E_{f}
\]
Assuming a Reference Level (R.L.) at the base of the inclined plane and the acceleration due to gravity as g. We have that at the top of the plane there is only potential energy due to height, kinetic energy is null, v0 = 0, at the base of the plane there is kinetic energy, the potential energy is null, h = 0.
Potential Energy is given by
\[ \bbox[#99CCFF,10px]
{U=mgH}
\]
Kinetic Energy is given by
\[ \bbox[#99CCFF,10px]
{K=\frac{mv^{2}}{2}}
\]
\[
\begin{gather}
U_{i}=K_{f}\\
mgH=\frac{mv^{2}}{2}\\
mgnk=\frac{mv_{T}^{2}}{2}\\
gnk=\frac{v_{T}^{2}}{2}\\
v_{T}^{2}=gnk \tag{VI}
\end{gather}
\]
substituting the expression (VI) into expression (V), the component in the r direction will be
\[
\begin{gather}
a_{r}=2gnk\frac{4\pi ^{2}R}{k^{2}+4\pi^{2}R^{2}}\\
a_{r}=\frac{8\pi ^{2}Rgnk}{k^{2}+4\pi^{2}R^{2}} \tag{VII}
\end{gather}
\]
The acceleration due to gravity points vertically down, the projection of acceleration due to gravity in the
direction of the inclined plane will be g sin α (Figure 5-A)
In the expression (I), the terms of the acceleration in the z and θ directions represent the vector acceleration tangent to the curve aT. This vector matches the component of acceleration due to gravity in the inclined plane (Figure 5-B)
\[
a_{T}=g\operatorname{sen}\alpha
\]
Note: The acceleration vector aT is tangent to the trajectory at
each point of the helicoid,
\( {\mathbf{a}}_{T}=a_{T}{\mathbf{e}}_{T} \),
where eT is the unit vector that gives the direction of the acceleration vector,
\( {\mathbf{e}}_{T}=\dfrac{{\mathbf{a}}_{T}}{a_{T}} \).
The vector acceleration can be written in terms of the unit vectors ez and eθ as \( {\mathbf{a}}_{T}=a_{\theta }{\mathbf{e}}_{\theta}-a_{z}{\mathbf{e}}_{z} \)
The vector acceleration can be written in terms of the unit vectors ez and eθ as \( {\mathbf{a}}_{T}=a_{\theta }{\mathbf{e}}_{\theta}-a_{z}{\mathbf{e}}_{z} \)
\[
{\mathbf{e}_{T}=\frac{a_{\theta}{\mathbf{e}}_{\theta}-a_{z}{\mathbf{e}}_{z}}{a_{T}}}
\]
Using the expression for the sine obtained in (IV)
\[
\begin{gather}
a_{T}=g\frac{k}{\sqrt{k^{2}+4\pi ^{2}R^{2}\;}} \tag{VIII}
\end{gather}
\]
The acceleration vector tangent to the trajectory can be written in terms of the components in directions
z and θ
\[
{\mathbf{a}}_{T}=a_{\theta}{\mathbf{e}}_{\theta}-a_{z}{\mathbf{e}}_{z}
\]
and the expression (I) may be rewritten as
\[
\begin{gather}
\mathbf{a}=-a_{r}{\mathbf{e}}_{r}+a_{T}{\mathbf{e}}_{T} \tag{IX}
\end{gather}
\]
substituting expressions (VII) and (VIII) into expression (IX)
\[
\mathbf{a}=-{\frac{8\pi^{2}Rgnk}{k^{2}+4\pi^{2}R^{2}}}{\mathbf{e}}_{r}+\frac{gk}{\sqrt{k^{2}+4\pi^{2}R^{2}\;}}{\mathbf{e}}_{T}
\]
factoring the term
\( \dfrac{gk}{\sqrt{k^{2}+4\pi ^{2}R^{2}\;}} \)
\[ \bbox[#FFCCCC,10px]
{\mathbf{a}=\frac{gk}{\sqrt{k^{2}+4\pi^{2}R^{2}\;}}\left(\frac{-{8\pi ^{2}Rn}}{\sqrt{\;k^{2}+4\pi^{2}R^{2}\;}}{\mathbf{e}}_{r}+{\mathbf{e}}_{T}\right)}
\]
and your module will be
\[
a=\frac{gk}{\sqrt{\;k^{2}+4\pi ^{2}R^{2}\;}}\left[\left(\frac{-{8\pi^{2}Rn}}{\sqrt{k^{2}+4\pi^{2}R^{2}\;}}\right)^{2}+1^{2}\right]^{1/2}
\]
\[ \bbox[#FFCCCC,10px]
{a=\frac{gk}{\sqrt{k^{2}+4\pi ^{2}R^{2}\;}}\left[\frac{64\pi^{4}R^{2}n^{2}}{k^{2}+4\pi ^{2}R^{2}\;}+1\right]^{1/2}}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .