What is Kinematics: Definition and 1000 Discussions
Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move. Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of mathematics. A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system. Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined. The study of how forces act on bodies falls within kinetics, not kinematics. For further details, see analytical dynamics.
Kinematics is used in astrophysics to describe the motion of celestial bodies and collections of such bodies. In mechanical engineering, robotics, and biomechanics kinematics is used to describe the motion of systems composed of joined parts (multi-link systems) such as an engine, a robotic arm or the human skeleton.
Geometric transformations, also called rigid transformations, are used to describe the movement of components in a mechanical system, simplifying the derivation of the equations of motion. They are also central to dynamic analysis.
Kinematic analysis is the process of measuring the kinematic quantities used to describe motion. In engineering, for instance, kinematic analysis may be used to find the range of movement for a given mechanism and working in reverse, using kinematic synthesis to design a mechanism for a desired range of motion. In addition, kinematics applies algebraic geometry to the study of the mechanical advantage of a mechanical system or mechanism.
I am confused because the question implies that I need to do some sort of calculation with Kepler's law. I got
##r+d = \sqrt[3]{\frac{T^2 GM}{4 \pi^2} } ##
But don't understand why I need this, since I already have the distance and the angular diameter should be ##\arctan (2R/d)## I think I...
I solved it using parallelogram law if vector addition but didn't got the correct answer.why?
Is their any other way to add velocity vectors.
How to do this problem
Hi, I have just joined the forum. Thank you all for being a part of such places so that people like me can get answers to the questions on their minds!
---------------------------
I have been trying to understand how a quadcopter yaws. Referring to the figure below which is bird's eye view of...
Hi folks,
See below for a solved question finding the down slope distance of an arrow. How easy would it be to solve this question by making the x-axis the slope direction?
I tried doing (vi)(∆ t) / 1/2(a)(∆ t) = 1/2(a)( ∆ t)² / 1/2(a)(∆ t)
That allows you to cross out the two accelerations out and you end with an answer of 20. But apparently, they never meet. I then tried making them into two different equations and using graphing technology (Desmos) and they...
Hi everyone
I have an issue with my physics teacher about a question on displacement. I have inserted the screenshot of the question below. By the definition of displacement, I was sure, and still am sure my answer is correct. But my physics teacher insisted that I am wrong. Please, can anyone...
Here is my attempt at the solution but I got the wrong answer. The right answer is t=16.6s. I know from the book (this is an example problem) that the motorcycle reaches its max speed at t=7.0s. But I don’t know where I made the mistake that is causing me to get the wrong answer afterwords.
Question is extracted from "Ellad B Tadmor, Ronald E Miller, Ryan S Elliott - Continuum mechanics and thermodynamics From fundamental concepts to governing equations".
I just got stuck at part (a). I think if part(a) is solved, I may be able to do the other parts.
5 links (thus 6 frames) and corresponding theta, alpha, d and a values are all given.
However, I'm not sure how to start coding the DH parameters to derive a forward kinematics model.
I know that I'll have to use matrix operators, but do I just put in the values of the 4 factors mentioned above...
We know that v=v(x, t) i.e. V is a function of time as well as space coordinate here I have only taken 1D motion for simplicity. By chain rule a=dv/dt= Dv/Dx*dx/dt +Dv/Dt*dt/dt where D /Dx represent partial differentiation along x coordinate.
So, a=Dv/Dx*dx/dt +Dv/Dt
Here we say that Dv/Dt=0...
A potter's wheel with a 35.9 cm radius rotates with a 2.91 rad/s2 angular acceleration. After 5.37 s, the wheel has rotated through an angle of 77.7 rad.
a)What linear distance did a point on the outer edge travel during the 5.37 s?
b)What was the initial angular velocity of the wheel?
c)What...
So, I recognise that:
$$ma=pg\left( L^{2}\right) \left( L-y\right) -\dfrac {1}{2}pgL^{3} $$
whereby $$pg\left( L^{2}\right) \left( L-y\right)$$ is the upthrust while the other is mg.
So, to find the largest speed of cube, I will assume that acceleration is zero since the acceleration slowly...
initial total KE= (1/2)(0.6kg)(8m/s)^2 = 19.2J
(0.6kg)(8m/s) = (0.6kg+1.8kg)(vf)
vf= 2m/s
final KE= (1/2)(0.6kg+1.8kg)(2m/s)^2 = 4.8J
I tried to use linear speed=angular speed * radius : thus
2m/s= angular speed * (3.3m/2)
angular speed= 1.2 rad/s
Apparently that is wrong.
I've been attempting to solve this problem for three days now. I have thrown away my old attempts (like, scrumpled up into the bin), but my old attempts involved:
Trying to set up simultaeneous equations relating the journeys between EH and FG to find the deceleration, but the reason why this...
Hello,
I've a small project (stewart platfrom) part of another one. I'd like to move my platform with a joystick on X, Y translations and roll, pitch, yaw.
Since this is a component of a bigger project, so this part should just move without precise kinematics, I'd like to perform approximate...
Hints given:
-Start with free body diagram. Use the relationship between impulse and momentum to find the final velocity of the car after he has pushed for time t.
-Use a kinematic equation to relate the final velocity and time to the distance traveled.
-What is his initial velocity?
My...
moment of inertia = (1/3) (2.1kg) (1.2m)^2 = 1.0 kgm^2
center of mass= (0.6i, 0j)
magnitude of the gravitational torque=9.8m/s^2*2.1kg*0.6m= 12.34N*m
position of the new center of mass now :
x direction = cos(20)*0.6m=0.56m
y direction= -sin(20) * 0.6m = -0.2m
change in gravitational...
For this problem I tried to find when the binoculars reaches the maximum height. So, (0.75 + 1.28)/2 = 1.015s. Using that information I can solve for the initial velocity. v_o = gt = 9.947 m/s.
Then using the initial velocity I can solve for the height of John using the 2nd kinematics equation...
I am new to physics and these are my last two questions of my homework. I am extremely confused on what to do but I do know that for question 1 I need to create a triangle and solve for the hypotenuse, and for question 2 I believe I need to input my given values into the equation for...
Hi, I tried to understand kinematics after having an omnidirectional roobt. Some problems stop me to go further. Here I upload some contents of different papers talking about kinematics. For the 1st three pictures, I don't know how equation (8) is from and I am little confused about...
Summary: Olympic problem from kinematics
Hello,
could anyone help me with the following problem? I don't quite get how exactly does it work.
After being kicked by a footballer, a ball started to fly straight towards the goal at velocity v = 25m/s making an angle α = arccos 0.8 with the...
Homework Statement: A box of mass M=30 kg sits on a ramp which is tilted at angle 13 degrees. The coefficients of static and kinetic friction between the box and the ramp are both 0.12. The box is connected to a spring of spring constant k=9 Newtons per meter.
Initially, the box is held in...
Homework Statement: Joe puts on his skis and heads for the slopes. A fresh layer of new snow has fallen, making the coefficient of kinetic friction between his skies and the snow only μk=0.037. He heads for the Starlight Run, which features a long slope at a constant angle of θ=17.8 degrees...
0.25g converts into 2.45m/s^2
V(t)=Vo+AoT
1=0+2.45T
T=1/2.45
T=0.4s
X(t)=VoT+1/2AoT
X(t)=100+0+1/2(2.45)(0.4)^2
X(t)=100.196m
I don't know if this is the right methodology or how to move on from here
So my problem isn't actually finding the components, but knowing if the initial approach I took is correct. So what I did was:
At first I found that at the same instant, ##x_{B/A}=10500 m## so then I wrote the equation of motion for plane B respect to A:
so $$\vec a_{B/O}- \vec a_{A/O}=\vec...
My working:
##s=\int v##
##v= \sqrt{\frac{a_{c}}{r}}=\sqrt{\frac{a_{c}}{\frac{4}{2t+2}}}##
##s= \int_{0}^{2} \sqrt{\frac{2}{\frac{4}{2t+2}}}##
My final answer seems to be wrong. Any ideas? Cheers
Well, what I've done so far is calculating the magnitude of velocity and acceleration replacing ##t=2## in ##\theta (t)## and ##r(t)## so I could get the expressions for ##\dot r##, ##\dot \theta##, ##\ddot r## and ##\ddot \theta##. But that's not my problem... my problem is related to the...
So I know that ##a_t = \frac{dv}{dt}=-ks## and ##\frac{dv}{dt}=v\frac{dv}{ds}## then: $$v dv=-ks ds \rightarrow (v(s))^2=-ks^2+c$$ and using my initial conditions it follows that: $$(3.6)^2=c \approx 13$$ and $$(1.8)^2=13-5.4k \rightarrow k=1.8 \rightarrow (v(s))^2=13-1.8s$$
What bothers me is...
The statement "at the initial moment of time v ⊥ u and the points are separated by a distance l " gives us a picture like the one which I have added in attachment.
As the time passes velocity vector v would gradually change from fully vertical to fully horizontal in order to meet point B. Now...
I've calculated the potential energy at the top of the halfpipe, before the boarder drops in:
PE = 39.5 kg * 9.8 m/s^2 * 3.66 m = 1416 J
Since the boarder would have no potential energy and all kinetic energy at the bottom of the halfpipe,
KE = 1/2mv^2 = 1416 J
1/2 (39.5 kg) (v^2) = 1416 J
So...
The car in front of us (2.car): ## s_2=v_0t ##
The car that is accelerating (1.car): ## s_1=x+2l+s_2= \frac{1}{2} at^2 ##
Now, if we substitute the equations, we get ## x+ 2l+ v_0t= \frac{1}{2} at^2 ##.
I have now 1equation with 2variables (a, t)- any suggestions on how to continue?
A particle, P, starts from rest at a point X and moves in a straight line with an acceleration expressed as a=4t. After 2 seconds, the particle reaches Y and it stops accelerating. The particle leaves Y with a velocity -3ms-1, and finally comes to rest at Z.
(i) Find the value of t when the...
My approach:
Let us take two orthogonal axes: x, parallel to the racket's plane and y, perpendicular to it. For the ball to not spin, the components of initial velocities of the racket and the ball along x-axis must be same. Also, as the line of collision is along the normal to the racket's...
I have a problem where I am given a trajectory by x(t), y(t) and I am given a constant speed throught the whole trajectory. I need to find vx and vy.
Equations that I am given:
x(t) = 1.5 + 0.5 * t * cos(8*pi*t)
y(t) = 1.5 + 0.5 * t * sin(8*pi*t)
v0 = 0.5
What I have tried to do is use the 1D...
Summary: Mechanics problem related with Calculus (differential equations)
Hi everyone, I would like some help in that task, if anyone would be willing to help :) Namely I have a problem from particle dynamics. "D:" means given info... so, D: m,g,h,b, miu. We're looking for v0 and S as given...
Step 1. Integrate the function --> it comes out to -1.8t3+t2-8t+c
Step 2. Figure out the integration constant --> I know that the velocity of the car is 30m/s at t=0s so making 30m/s = -1.8(0s)3+(0s)2-8(0s)+C it would seem that C is 30m/s
Step 3. Calculate velocity at 4s --> V(4s)=...
Note: Maths has always been kinda a weak point for me in school. Anyway.
From hereon I'm going to talk about this in the context of a train for simplicity, even though this is actually needed for an entirely different context - but I don't think this is relevant for the problem.
In terms of my...
I selected A, because the ball is merely in motion due to the force of gravity (after applied force).
My teacher has not concerned us with air resistance so can someone please explain why the correct answer is C? I reasoned that gravity was always accelerating the ball at -9.8 m/s^2...