Kinematics Definition and 1000 Threads

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...

Vx = 10cos(23) = 9.21m/s Vy = 10sin(23) = 3.91m/s -30 = 3.91t + 1/2(-9.8)t^2 0 = t(3.91 - 4.9t) + 30 4.9t = 33.91 t = 6.92s Delta X = 9.21(6.92) + 1/2(9.8)(6.92)^2 Delta X = 298.38m Vf^2 = 9.21 + 2(9.8)(298.38) Vf^2 = 5857.458 Vf = 76.53m/s

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...

I have been attempting this physics problem for a while now. I am not sure how to begin to attempt this problem with the equations of motion formulas.

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...

So I worked out the first part and obtained ##E_1 = 478.8MeV##, ##E_2 = 459.4MeV## and ##p = 0.49 MeV/c## but I can't quite wrap my head around the second part. Normally, I'd use the equation for s but I'm confused since I don't know the angle between the gamma rays.

So above is the question. The picture below is my attempt at the solution: However the answer at the back of the book suggest that T=1.929s

What I attempted to do first was find alpha and turn that into translational acceleration. Taking mass of yoyo * radius of spool * gravity, (3kg)(0.8m)(9.81m/s^2) yielded 23.544 N*m, and dividing by I = 7.68 kg * m^2 yielded 3.065625 rad/s^2. Finally, multiplying by r = 0.8m gave me 2.4525...

im confused

Hi I am trying to learn kinetics and kinematics: Euler angles Directional cosine matrices Quaternions Inertia tensors The usage of Jacobians in robotics Manipulator kinematics Eulers equations Newton-Euler equations etc My background is in mathematics, so I would like a book which takes a...

Homework Statement The graph shows the variation with time t of the acceleration a of an object (graph attached). What is the change in velocity of the object in the time interval 0 to 4s? Homework Equations a = v/t The Attempt at a Solution My initial answer was -8 m/s, by simply finding the...

Homework Statement Hello. My colleague and I have been struggling with this assignment where we have to: - Given the position and orientation of the end-effector of a robotic arm with 6 degrees of freedom get the joint angle values (inverse kinematics). The robotic arm in question is the...

Hi all, The scenario I'm considering is a solid sphere (of uniform density) rotating with constant angular velocity when it abruptly splits into two hemispheres along a cut which contains the rotation axis. The hemispheres will begin to separate; if, for example, we consider the rotation to be...

Homework Statement This is a physics problem that I made up in order to understand one aspect of kinematics in 1-dimension: A baseball is thrown straight up, leaving the hand which is 1.1 meter above the ground. It reaches a height of 12 meters. (a) What must be the speed of the baseball when...

Homework Statement A person of mass ##m## stands at the left end of a boat of mass ##M## and length ##L##. Both the person and the boat are initially at rest. The coefficient of kinetic friction between the boat and the water is ##\mu##. If the person starts moving to the right, what is the...

Hi I am learning kinematics. Topics include inertia tensors, cosine matrices, quaternions, Euler angles etc... To learn these topics well, I want to try working on something a bit more difficult than just the underlying math. That will probably keep me more motivated. So, I am looking for...

So the kinematics of the contact point of a disc rolling without slip in a Cartesian plane if fairly straightforward. The velocity for the contact point is just v = ω r ev where r is the wheel radius and ev the current direction vector of the wheel. However, say that you grab the plate and...

Homework Statement A honeybee is flying parallel to tabletop at height with velocity .With its wings it can acquire a maximum acceleration of . At an instant when honeybee is vertically above honey drop on tabletop and decides to reach the honey drop. Neglect reaction time of honeybee and...

Here's the problem: https://i.stack.imgur.com/rUVvu.png What I did: [2]: https://i.stack.imgur.com/AX2Ye.png [3]: https://i.stack.imgur.com/K1Zbi.png To those who could not understand what I read: So basically I used some geometry to to get $l_1$ and $l_2$ in terms of $r$ and the assumed...

Homework Statement First problem: The speedometer in a car is driven by a cable connected to the shaft that turns the car's wheels. Will speedometer readings be more or less than actual speed when the car's whells are replaced with smaller ones? Second problem: Keeping in mind the concpet from...

Homework Statement Homework Equations Vy2 = Vyo2 + 2ayt y = 1/2(Vyo + Vy)t The Attempt at a Solution So I have "solved" this problem with both the equations above. Using the Vy2 equation I get the result of 32.83m/s. Using the other equation I get a result of zero, because Δy = (0 - 55) =...

Homework Statement A passenger train is traveling at 29ms^-1 and a freight train is traveling 360m ahead of it at 6ms^-1 in the same direction on the same track. The driver of the passenger train has a reaction time of 0.4s before he starts decelerating. What is the minimum deceleration to...