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Project 1a
Rectilinear Kinematics
Rectilinear kinematics is study of bodies/particles moving in straight line.

Four Differential Relationships
The four differential relationships that governs the velocity and acceleration of particles moving along straight line is given below
v=dsdta= dvdta= d2sdt2a= vdvdsWhere s is the position of the particle along the line, ? is the particle’s velocity, and ? is the acceleration of the particle.

We use these equations to calculate the velocity and acceleration of particles moving along straight line.

Constant Acceleration Equations
The equations for a particle moving with constant acceleration along straight line are
v=vo+acts=so+ vot+act22v2=vo2+2ac(s-so)These equations are only applied when a particle is moving with constant acceleration.

Difference Between Speed And Velocity
The difference between speed and velocity is that speed is a scalar quantity while velocity is a vector quantity.
Difference Between Average And Instantaneous Velocity
Average velocity gives us the average magnitude at different instants while instantaneous velocity is a velocity at that particular instant.

Problem Solution:

Figure 1: The planar manipulator is constrained to only move in the horizontal direction.

As it is given that, the acceleration of the end-point of the manipulator will be controlled by the mathematical relationship r=kt ms2 ; where k is constant.

It can be assumed that the manipulator starts from rest at ? = 1 m. The main purpose of solving this numerical problem is to find the value of k, when manipulator reaches r= 2 m after 3 seconds.

Let’s assume manipulator starts at t= 0 sec
r= drdt=ktrordr= otktdt Integrating and applying limits we get;
r=k2t2+roms (1)We will get equation for position vector “r” by integrating equation (1)
r=drdt=k2(t2-t02)+roAs t0= 0 sec
dr=k2t2+rodtrordr=k2t2+rodt (2)Integrating and applying limits we get;
r= k6t3+rot+ro mSolving for k; we get
k=6(r-ro-rot)t3 (3)Putting values r= 2 m, ro = 1 m , ro = 0 (starting from rest) and t = 3 sec
k=6(2-1-0*3)33k=29 ms3
Figure 2: A plot the position, velocity, and acceleration of the end-point of the machine as time varies from ?= 0 to ? = 3 seconds.

Project 1b
General Curvilinear Motion:
“A particle is said to be moving in curvilinear motion if travels along curved path”
In rectilinear motion a particle always moves along straight path while in case of curvilinear motion the particle always travels along curved path.

Secondly, in rectilinear motion the velocity of particles is uniform while in curvilinear motion particle velocity changes at every point due to change in direction.
Rectangular Coordinate System:
The rectangular coordinate system consists of two real number lines that intersect at a right angle. The horizontal number line is called the x-axis, and the vertical number line is called the y-axis. It is used for the demonstration of various 2D geometrical and non-geometrical shapes.

The origin of rectangular coordinate system is placed at point “0” in both axes.

The coordinate axes are said to be decoupled, if they transact without being connected.

Figure.1: Rectangular (Cartesian) coordinate system with the origin at the centre of the revolute joint.

Position Vector:

Figure 2: Conversion of polar coordinates into rectangular (Cartesian) coordinate system with the origin at the centre of the revolute joint.

According to paythagorous theorem;
From figure 2;
cos?=xr and sin?= yrSo x=rcos? and y=rsin?So position vector;
r=xcos?i+ysin?j (1)Velocity:
3609109117752140997331089141?00?3788468673851er00er2867891542404e?00e?
Figure 3: Conversion of polar coordinates into rectangular (Cartesian) coordinate system with the origin at the centre of the revolute joint along with unit vectors.

As we know in polar coordinates velocity;
v= rer+r?e? (2)From figure 3;
er=cos?i+sin?je?= -sin?i+cos?jPutting in above equation (2):
v= r(cos?i+sin?j)+r?(-sin?i+cos?j)Re-arranging
v=(rcos?i -r?sin?i)+(rsin?j+r?cos?j)v=(rcos?-r?sin?)i +(rsin?+r?cos?)Acceleration:
As we know in polar coordinates for acceleration;
a=r-r?2er+r?+2r?e? (3)As we know from figure 3;
er=cos?i+sin?je?= -sin?i+cos?jPutting in the equations (3)
a=(r-r?2)(cos?i+sin?j)+(r?+2r?)(-sin?i+cos?j)a=(rcos?i-r?2cos?i+rsin?j-r?2sin?j)+(-r?sin?i-2r?sin?i+r?cos?j+2r?cos?j)Rearranging;
a=(rcos?i-r?2cos?i-r?sin?i-2r?sin?i)+(rsin?j-r?2sin?j+r?cos?j+2r?cos?j)a=(rcos?-r?2cos?-r?sin?-2r?sin?)i+(rsin?-r?2sin?+r?cos?+2r?cos?)jMagnitude Of The Position Vector
From equation 1;
r=xcos?i+ysin?jr=?(rcos?)2+(rsin?)2Putting values r= 4 m and ?=55°r=?(4cos55°)2+(4sin55°)2r=?(5.26)+(10.74)r=4 mMagnitude Of The Speed
From equation 2;
v=?(rcos?-r?sin?)2 +(rsin?+r?cos?)2Putting values = 55º, ? = 4 m, ?= 5 rad/s and ?? = 10 m /s
v=?(10*cos55°-4*5*sin55°)2 +(10*sin55°+4*5*cos55°)2v=?(113.27) +(386.576)v=?(500)v=22.36 m/sMagnitude Of The Acceleration:
From equation 3;
a= ?(rcos?-r?2cos?-r?sin?-2r?sin?)2+ (rsin?-r?2sin?+r?cos?+2r?cos?)2Putting values:
? = 55º, ? = 4 m, ?= 5 rad/s and ?? = 10 m /s, ?=5 rad/s2 and r=10 m/s2a= ?(10*cos55°-4*52*cos55°-4*5*sin55°-2*10*5*sin55°)2+ (10*sin55°-4*52*sin55°+4*5*cos55°+2*10*5*cos55°)2a= ?22481+24a=150 m2s