# Tag Info

29

You can treat "just before" and "just after" as left-hand limits and right-hand limits respectively. If you want the behavior of some quantity that is a function of time $f(t)$, then $f$ right before and right after $t=t_0$ would be $$f_\text{before}=\lim_{t\to t_0^-}f(t)$$ $$f_\text{after}=\lim_{t\to t_0^+}f(t)$$ Other answers go into ...

17

Jerk is the change in acceleration. If acceleration is changing, then there is jerk. Since there is a changing acceleration in circular motion, there is jerk in circular motion. I'm not entirely sure what you mean by "manifest" though. It's simply something that can be measured.

8

As other answers and comments said, yes there is jerk. But it doesn't mean what you think. In everyday use, a jerk is a sudden acceleration, like a sudden yank on a rope. That is not what centripetal acceleration is like. If you tie a rope to a rock and swing it around your head, you must pull on the rope to keep it moving in a circle. That is centripetal ...

8

I am sure that you have met this idea before but not noticed it. For example when you are investigating the collision of two objects you most often deal with before the collision and after collision but never even consider what is happening during the collision and the time during which the collision occurs. In such an example the approximation is made that ...

7

I actually gave this as a project to my students to do with Super Mario. The trick is to use this fantastic software called Tracker Video Analysis which allows you to analyse video files to find the position, velocity, and acceleration of different objects. You'll need some scale of reference to be able to do this (I believe my students used Mario's height ...

6

In uniform circular motion, the velocity has a constant magnitude. In this case the acceleration always points towards the center of the circle (perpendicular to the velocity) and has a magnitude exactly equal to $v^2/r$. In the more general sense, we can have acceleration that has a tangential component that points along the circle (parallel/anti-parallel ...

6

$K.E=\frac{1}{2}mv^2$, $P=mv$ thus there is a relation between them. I really cant understand why there is no decrease in momentum when kinetic energy is decreased in inelastic collision. It doesn't make sense to consider a single body during a collision (i.e. during interaction with a second body). You need to look at both bodies together. Therefore you ...

5

I think there is a teachable moment here, in that I think we can get some insight through a different way to frame your question. The first thing to note is that $\omega$ has dimensions of $[{\rm time}]^{-1}$ (in SI units, ${\rm s}^{-1}$). Therefore, it does not make sense to compare $\omega$ to a dimensionless number, like $1$, so saying $\omega>1$ or $\... 5 If a body orbits something else, then there is gravitational waves (GW) emission and the orbit shrinks (eventually the two bodies will merge). Neglecting GW emission (assuming Newtonian dynamics is the correct description), the motion goes on forever, but this is not problematic since any kind of friction is neglected. So the system is no more or less ... 5 So you aren't doing the same thing as your teacher here. Notice how in what your teacher did they have$\Vert d \mathbf r \Vert =ds$, so they did not say that $$\frac{d\mathbf r}{dt}·\frac{d\mathbf r}{dt} = \left(\frac {dr}{dt}\right)^2$$ this would not be true in say circular motion, where the position vector is changing but its magnitude is not. However, ... 4 To start off, it seems like you are thinking about uniform circular motion, as you are fixated on the centripetal acceleration and are not mentioning anything about tangential acceleration. Therefore, for now let's assume we are talking purely about uniform circular motion. For circular motion to happen we need a centripetal force or acceleration i.e. a ... 4 Just to add a bit of more mathematical flavour to the answers, here's why momentum can stay constant, yet kinetic energy may decrease. First we set up some ground rules. You mention that momentum is$p=mv$and kinetic energy is$\frac12mv^2$and that there should be a relation between them. And yes, there is. Basic algebra tells you$K=\frac{p^2}{2m}$. Well ... 4 As for measuring the rebound height, I would suggest you use a camera (make sure the camera is fixed, not hand held). Drop the ball alongside a wall with a ruler or other tape measure fixed on the wall. You may get errors due to parallax so place the camera relative to the ball and ruler so as to minimise errors caused by parallax. Note the height of the ... 3 Jerk is the change in acceleration, and in your scenario acceleration is constant and therefore jerk is zero. But no, because now we are considering the change in gravity due to height. So at any point, the acceleration is a function of height$h$only. $$a= \frac{F}{m} - \frac{GM}{(R+h)^2} \tag{1}$$ Considering the velocity as$v = \tfrac{\rm d}{{\rm d}t} ...

3

Although, I cannot comment on the specifics of your problem (due to the homework policy and the lack of details in the question), I can make a general statement of what it means by "before" and "after". In physics you are asked to create mathematics models that correspond to particular situations. Sometimes these situations undergo a ...

3

Yes, $\vec v=\frac{ds}{dt}\hat{t}$ So, $$d\vec{v}\cdot d\vec {v}=(d^2s\hat t +ds\,d\hat t)\cdot(d^2s\hat t+ds\,d\hat t)=(d^2s)^2+(ds)^2$$ But ${dv}^2=(d^2s)^2$

2

why turning the wheel of a static car is much harder than turning one in motion Straight answer is that in a static car you are overcoming static friction and when in motion - rolling resistance. Static friction coefficient for tires is about $50\times$ higher than a rolling resistance coefficient for same tires.

2

A net force or torque on a rigid body will not affect its internal energy. As it remains constant before and after the application of the force/torque, it is not relevant to the equations of motion.

2

I want to address what you said about the aim being to "complete his circular motion". I assume you mean that the particle is moving in a circle with a constant radius. Let $v$ be the speed of the particle, $a_{||}$ be the component of the acceleration in the direction of the velocity of the particle, and $a_{\perp}$ be the component of the ...

2

The velocity vector describes the change in the displacement vector, and hence, it can be in any direction relative to what direction the displacement vector points in, since there are many ways the position of the particle can change. A few counterexamples: Circular motion: Velocity vector is perpendicular to the position vector which denotes the particle ...

2

Displacement is the shortest possible distance between the initial point and final point of line of motion. Do velocity and displacement always have the same direction? Need not to be. Imagine a ball being projected vertically upward with some speed $v$. At some time t (less than time of ascent i.e. $t<\frac{v}{g}$ ), the direction of its velocity is ...

2

The answer for all the Why's ? Part (iii) : By struck , the question indicates an impulsive force acting on the block and giving speed to the block . So yes , It accelerates because of this force but only for the time the force exists. But the question asks for the distance travelled after it gains the speed of $5.5 m/sec$ i.e. the question starts after the ...

2

Actually there is no relationship between Kinetic Energy and Momentum (for a generalized system of particles). You can have a non-zero kinetic energy at the same time with momentum being zero. You can see this from the following equations: $$\mathcal E = \frac 12 \sum m_i v_i^2$$ $$\mathbf P = \sum m_i \mathbf v_i$$ Now if $v_i \neq 0$ then $\mathcal E$ will ...

1

Calculate their relative velocity. Determine how long it will take for both vehicles to meet over that distance. From there it will be trivial to calculate the distance the SUV travels.

1

those two wheels are rotating on the same axle, but at different radii. This means that if the axle makes one complete turn or revolution around one of its ends, the wheels will trace out different circumferences as they roll on the ground, and their angular velocities will be different. The innermost wheel will trace out a smaller circumference and rotate ...

1

Why should the tension force change ? Both the questions are talking about equilibrium condition and in both the question , $A$ is at rest and is about to move either downward or upward. So , tension in both the case is counter balancing weight of $A$(thus equals $2N$) For B : When the new Force is introduced , it has a perpendicular as well as a horizontal ...

1

In the second part of the question they are again hinting to the fact that body A is in equilibrium . That's why the tension is again $2 N$. The other hint that the body A is about to move upward points that friction acting on the body B is leftward.

1

Sketch C is drawn correctly for a glass of water under horizontal acceleration. As it is accelerating gravity is not the only force acting on it, so it would not remain horizontally level. Tilting the tray would be a way to keep the glass from spilling or sliding on the tray while accelerating. However, I doubt that very many waiters would accelerate quickly ...

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