Particle Kinematics

Particle kinematics is a branch of Classical mechanics that studies the geometrical propertries of motion of a particle (point mass)

### 1] Velocity and acceleration of a point particle

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The position of a point particle S in a reference frame is defined by a vector function  r → {\displaystyle {\vec {r}}} (t) from an arbitary point O.The set of all the points that specify the position of a particle is called trajectory of the body. In the image bellow, the trajectory of the point mass is a curve.
Note that, in this system, the vector function is  r → = r → ( t ) {\displaystyle {\vec {r}}={\vec {r}}(t)} . This form is independent from any coordinate system.

Image (1)

The velocity of (S) is defined as the first derivative of position with respect to time. Therefore,
v → = d r → d t = r → ˙ {\displaystyle {\vec {v}}={\frac {d{\vec {r}}}{dt}}={\dot {\vec {r}}}}

As it seems, velocity is a vector quantity always tangential to the path of the object.

The acceleration of (S) is defined as the first derivative of velocity with respect to time. Therefore,
a → = d v → d t = v → ˙ = d 2 r → d t 2 = r → ¨ {\displaystyle {\vec {a}}={\frac {d{\vec {v}}}{dt}}={\dot {\vec {v}}}={\frac {d^{2}{\vec {r}}}{dt^{2}}}={\ddot {\vec {r}}}}

### 2] Centripetal and tangential acceleration

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In the previous paragraph,the independent variable of position was time, but we can hypothesize that the vector function  r → {\displaystyle {\vec {r}}} is a function of the distance s (see Image 1) along the trajectory. This can be expressed as:
r → = r → ( s ) {\displaystyle {\vec {r}}={\vec {r}}(s)}
and
s = s ( t ) {\displaystyle s=s(t)}

In order to find the velocity we differentiate the position vector with respect to time.

If we differentiate the vector function r with respect to time, we get:

d r → d t = d r → d s d s d t = v ε ^ {\displaystyle {\frac {d{\vec {r}}}{dt}}={\frac {d{\vec {r}}}{ds}}{\frac {ds}{dt}}=v{\hat {\varepsilon }}}

where  ε ^ {\displaystyle {\hat {\varepsilon }}} is the tangential unit vector to the path and it is a vector function of time.

In order to find the acceleration of the particle (S) we differentiate velocity with respect to time. Note that  s ˙ = v = v ( t ) {\displaystyle {\dot {s}}=v=v(t)} and  ε ^ ( t ) {\displaystyle {\hat {\varepsilon }}(t)} , that means that both speed (scalar quantity) and the unit vector are functions of time. So, using the product rule, we get:

a → = d v d t ε ^ + v d ε ^ d t ( 1 ) {\displaystyle {\vec {a}}={\frac {dv}{dt}}{\hat {\varepsilon }}+v{\frac {d{\hat {\varepsilon }}}{dt}}\quad (1)} But it is true that:

d ε ^ d t = d ε ^ d s d s d t = d ε ^ d s s ˙ {\displaystyle {\frac {d{\hat {\varepsilon }}}{dt}}={\frac {d{\hat {\varepsilon }}}{ds}}{\frac {ds}{dt}}={\frac {d{\hat {\varepsilon }}}{ds}}{\dot {s}}} Appying this to (1), we get

a → = d v d t ε ^ + d ε ^ d s v 2 ( 1.1 ) {\displaystyle {\vec {a}}={\frac {dv}{dt}}{\hat {\varepsilon }}+{\frac {d{\hat {\varepsilon }}}{ds}}v^{2}\quad (1.1)} Taking into consideration Frenet-Serret formulas, we know that  d ε ^ d s = ν ^ R {\displaystyle {\frac {d{\hat {\varepsilon }}}{ds}}={\frac {\hat {\nu }}{R}}} where ν is the unit vector perpendicular to the path.

So (1.1) becomes. a → = d v d t ε ^ + v 2 R ν ^ ( 1.3 ) {\displaystyle {\vec {a}}={\frac {dv}{dt}}{\hat {\varepsilon }}+{\frac {v^{2}}{R}}{\hat {\nu }}\quad (1.3)} We conclude that acceleration is analyzed in two perpendicular components. One in the direction of the path  ( d v d t ε ^ ) {\displaystyle ({\frac {dv}{dt}}{\hat {\varepsilon }})} and one perpendicular to the infinitesimal path

( v 2 R ν ^ ) {\displaystyle {\frac {v^{2}}{R}}{\hat {\nu }})} .The first component is called tangential acceleration and the second centripetal acceleration becuase its direction is pointing the center of curvature of the path.

Notice that when  s ˙ = 0 ⇒ a → = v 2 R n ^ ⇔ | a → | = v 2 R {\displaystyle {\dot {s}}=0\Rightarrow {\vec {a}}={\frac {v^{2}}{R}}{\hat {n}}\Leftrightarrow |{\vec {a}}|={\frac {v^{2}}{R}}} This case of motion is called uniform circular motion, where the trajectory of (S) is a circle of radius R and the magnitude of its velocity constant

### 3] Cartesian Coordinates

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In the previous paragraphs, the formulas were in vector form and thus, independent of any coordinate system. But in physics, it is useful to use coordinate sytems for the specification of the position of a particle in 3D space.

if (x,y,z) are the coordinates of a point mass S in 3D space then it is true that: S = ⟨ x ( t ) , y ( t ) , z ( t ) ⟩ {\displaystyle S=\langle x(t),y(t),z(t)\rangle } And then the position vector that specifies the position of S is:

O S → ( t ) = r → ( t ) = x ( t ) i ^ + y ( t ) j ^ + z ( t ) k ^ {\displaystyle {\overrightarrow {OS}}(t)={\vec {r}}(t)=x(t){\hat {i}}+y(t){\hat {j}}+z(t){\hat {k}}} where x(t), y(t) and z(t) the component functions the vector function r and i,j,k, the unit vectors that represent x,y and z axes respectively.

i ^ = [ 1 0 0 ] , j ^ = [ 0 1 0 ] , k ^ = [ 0 0 1 ] {\displaystyle {\hat {i}}={\begin{bmatrix}1\\\0\\\0\end{bmatrix}},{\hat {j}}={\begin{bmatrix}0\\\1\\\0\end{bmatrix}},{\hat {k}}={\begin{bmatrix}0\\\0\\\1\end{bmatrix}}}

The velocity of (S) is:

r → ˙ = x ˙ i ^ + y ˙ j ^ + z ˙ k ^ {\displaystyle {\dot {\vec {r}}}={\dot {x}}{\hat {i}}+{\dot {y}}{\hat {j}}+{\dot {z}}{\hat {k}}} and the acceleration is given by:

r → ¨ = x ¨ i ^ + y ¨ j ^ + z ¨ k ^ {\displaystyle {\ddot {\vec {r}}}={\ddot {x}}{\hat {i}}+{\ddot {y}}{\hat {j}}+{\ddot {z}}{\hat {k}}} Exercises

1] Find the velocities and accelerations of patricles moving with the position functions bellow. Which one of the vector functions can't be a position function and why?

a) r → = ⟨ t , 5 , − t 2 − 3 t − 2 ) ⟩ {\displaystyle {\vec {r}}=\langle t,5,{\sqrt {-t^{2}-3t-2)}}\rangle }

b) r → = ⟨ 3 t 2 , 5 t , 3 ⟩ {\displaystyle {\vec {r}}=\langle 3t^{2},5t,3\rangle }

c) r → = ⟨ t , t 3 − 10 t + 5 , 0 ⟩ {\displaystyle {\vec {r}}=\langle t,t^{3}-10t+5,0\rangle }

d) r → = ⟨ 3 t , 4 t 2 , − 10 ⟩ {\displaystyle {\vec {r}}=\langle 3t,4t^{2},-10\rangle }

e) r → = ⟨ t , e t , 1 ⟩ {\displaystyle {\vec {r}}=\langle t,e^{t},1\rangle }

f) r → = ⟨ 5 t , c o s x ( t ) ⟩ {\displaystyle {\vec {r}}=\langle 5t,cosx(t)\rangle }

g) r → = ⟨ t 6 , s i n ( 5 t ) , l n ( t + 2 ) ⟩ {\displaystyle {\vec {r}}=\langle t^{6},sin(5t),ln(t+2)\rangle }

h) r → = ⟨ t , 4 − ( t 2 + 3 t + 2 ) , l n ( t ) ⟩ {\displaystyle {\vec {r}}=\langle t,{\frac {4}{\sqrt {-(t^{2}+3t+2)}}},ln(t)\rangle }

i) r → = ⟨ t , 1 ⟩ {\displaystyle {\vec {r}}=\langle t,1\rangle }

j) r → = ⟨ e t , e t , 9 − | t + 5 | 2 ⟩ {\displaystyle {\vec {r}}=\langle e^{t},{\sqrt {e^{t}}},{\sqrt {9-|t+5|^{2}}}\rangle }

2]A body P is moving in space with a position function  r → = ⟨ 3 t , 7 t 2 , t ⟩ {\displaystyle {\vec {r}}=\langle 3t,7t^{2},t\rangle } . Find the components of velocity and acceleration in the x,y and z axis

3]A point mass G is moving along an elliptical trajectory. Prove that the acceleration of the particle G is a function of position.

4]A particle K has acceleration  a → = ⟨ 1 , 3 , 6 ⟩ {\displaystyle {\vec {a}}=\langle 1,3,6\rangle } .Find the velocity and the position functions of the particle K if  v → o = ⟨ 2 , 4 , 1 ⟩ a n d r → o = ⟨ 2 , 7 , 3 ⟩ {\displaystyle {\vec {v}}_{o}=\langle 2,4,1\rangle \quad and\quad {\vec {r}}_{o}=\langle 2,7,3\rangle } .

5]Point mass is moving on a plane with constant speed. Prove that the acceleration vector is perpendicular to the velocity vector.

6]A body is moving in the xy plane, with constant speed and costant magnitude of acceleration. Prove that the body's trajectory is a circle.

Πηγές: "Θεωρητική Μηχανική", Ιωάννης Δ. Χατζηδημητρίου, Τόμος Α', Εκδόσεις Γιαχούδη Θεσσαλονίκη 2000.

AndreasMastronikolis (discuss • contribs) 14:05, 27 December 2016