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14

In this case, gravity is still an external force. In a zero-g environment, the mouse would also begin to move around the inside of the wheel, opposite the rotation it causes in the wheel, which would keep the angular momentum at zero. This would happen because the only way for the mouse to exert a force on the wheel and rotate it is for it to push itself in ...


6

Your intuition is correct. For the ball to change its angular momentum (to go from "backspin" to "forward spin"), there needs to be a net torque acting. There are two forces on the ball: gravity, and the normal force of the slope. Both these forces act through the center of mass - so neither force adds torque. Without torque, there is no change in angular ...


4

If you were to hit yourself (to be specific, let's say you're using your hand to hit your chest) then your hand would indeed produce a force on your chest, but your chest would also produce a force on your hand that is equal in strength and opposite in direction (Newton's Third Law). The combined effect of these two forces (when looking at your body as a ...


3

Yes, there are an infinite number of solutions, though your teacher will want you to choose the most obvious one. When the force does work on the mass, that work can be converted into two forms: the potential energy of the object the kinetic energy of the object If you apply a force of $800g$ then once the object has been raised the 2.4m it will still ...


3

What matters here is how the value of $c$ compares to the value of $k$. Let us choose a $\zeta = \frac{c}{2\sqrt{mk}}$ One can show that when $\zeta =1 $ the system is critically damped, and will not exhibited any oscillations and will return to the origin in the shortest possible time interval. When $\zeta > 1 $ the system is over damped and will take ...


2

Equations of motions need not be proven: they are such because they are experimentally true and there is no basic reason for that (at this point I wonder how you proved that $\dot{\textbf{p}} = m \textbf{a}$: you must have done so just re-writing a different form of the same equation, or any other starting point which you assumed to hold true). In the ...


2

I had this kind of question myself for a long time because before studying Physics I've studied Math and in Mathematics we do things quite differently. In Mathematics, a general procedure is to give some definitions, probably by specifying some axioms, then we derive and prove theorems from this. On Physics there is a similar procedure, but it goes a little ...


2

The time derivative of $v^2$ is $2v \frac{dv}{dt}$ not $2v$. You must use the chain rule.


2

Here is the procedure: $KE = 0.5mv^2$ $\frac{d}{dt}KE = 0.5m\frac{d}{dt}v^2$ So the question becomes,how do we find the derivative of $v^2$ with respect to time? One can easily see that $\frac{d}{dt} = \frac{dv}{dt}\frac{d}{dv}$ (Notice how the $dv$ cancels top and bottom) Therefore, $\frac{d}{dt}v^2 = \frac{dv}{dt}\frac{d}{dv}v^2 = \frac{dv}{dt}\times ...


2

The black machine is a weight lifting machine. It is self contained with no power source. If it can lift an external weight and return to its original state as shown below, it is a perpetual motion machine. Suppose the blue weight is water. We could add a water wheel and generator on the right. You start at the top and work your way to the bottom. Then you ...


2

I suspect this is an example of the spinning-egg problem, in which a prolate spheroid (such as an egg) spun on a table about one of its "short" axes will tend to "stand up" so that it's spinning about its long axis. A few explanations have been proposed for this phenomenon, most notably: H. K. Moffatt & Y. Shimomura, "Spinning eggs — a paradox ...


2

The direction of the motion at any time $t$ is the direction of the velocity vector $\textbf{v}(t)$ as derived by solving the equations of motion; likewise $\omega(t)$ gives you back the direction of rotation according to the right hand rule. friction is the force that causes rotation is not entirely correct. Any force with non-zero torque generates ...


1

The angular momentum of the earth && mouse wheel system does not change. When the earth pulls on the mouse, the mouse pulls on the earth, so no net moment is seen any arbitrary point in the universe.


1

In a reference frame where the center of mass of the Earth-Sun system is at rest, we will have $$ m_\text{Sun} \vec{v}_\text{Sun} + m_\text{Earth} \vec{v}_\text{Earth} = 0 \quad \Rightarrow \quad \vec{v}_\text{Sun} = - \frac{m_\text{Earth}}{m_\text{Sun}} \vec{v}_\text{Earth}. $$ In particular, if this is true at some initial time, then it's true at all ...


1

I am going to assume that you have not yet studied linear algebra, sorry if it seems as if I am talking down to you at any point. You are correct in that we can split a vector into two components in the plane. This is because any two linearly independent(not parallel or anti-parallel) vectors form a basis(a set of vectors from which you can "build" other ...


1

Your last equation is a quadratic in $t$. The $a$ is simply $\sin(\theta).g$. You can then solve it with the usual formula for a quadratic equation. There are two solutions to a quadratic, and that's because if you go into negative time you'd be pulled down by gravity any get to the new X position. This solution, of course, wouldn't apply to your ...


1

This becomes easier to understand if you think of velocity being made up of perpendicular components. For example, let $v = v_x\hat i + v_y\hat j$. That is, velocity is made up of an x-component, $v_x$ and a y-component, $v_y$. When there is a change of direction, the $v_x$ and $v_y$ components will change. This means there must be some horizontal ...


1

But we also know that a perpendicular force always causes an acceleration according to the rule of addition of forces. All forces cause acceleration. Perhaps you mean specifically tangential acceleration (changes in speed)? If the centripetal force is greater the resulting vector is near the perpendicular, if the centripetal force is in perfect ...


1

Answer to the question in the title? Two vectors are only equal to each other if they are the same (this is a general rule: equality means the things compared are identical). That means having the same direction as well as the same magnitude. So how could changing the direction of motion not be acceleration? Don't get hung up on fact that in 1 dimension ...


1

Is this the correct way to find the derivative of kinetic energy? $$ K=\frac{1}{2}m v^2 \\ $$ So: $$ \frac{dK}{dt} = \frac{1}{2} \left(\frac{dm}{dt} v^2 + 2mv \frac{dv}{dt} \right) $$ If the mass does not change over the time, then $$\frac{dm}{dt}=0$$ And finally $$ \frac{dK}{dt} = \frac{1}{2} \left(2mv \frac{dv}{dt} \right) $$ So simplifying: $$ ...


1

What data do you have for linear motion? Your equations are correct if you have the acceleration as a function of time and the orientation is constant. The angular accelerometer can give you the angles as a function of time with integration. Unfortunately, drift can be a problem. The received wisdom is to use an accelerometer (linear or angle), integrate ...


1

Let me recall what Newton's laws are, to start with: 1) In the universe exists at least one reference frame (that we call inertial) where $\textbf{v}= \textrm{const.}$ whenever no external interactions act on the particle. All other reference frames (if any) moving at constant speed wrt this very one will be inertial as well. 2) In the above reference ...


1

The usual approach: write down conservation of angular momentum, linear momentum, and energy. Assume the impact is elastic and infinitely short duration. In that time the spring didn't move and the third particle didn't come into the equation. That means the problem can be reduced to two simpler problems: two particles that hit elastically (after collision ...


1

There is deceleration caused by friction and drag, which is complicated as far as computing is concerned (as someone noted above). It can be determined empirically with help of some controlled experiments and curve fitting. The simplest is off course to assume that it is constant deceleration. Depending on the application, it may suffice. If so, the ...


1

You are correct, they are different. $k$ is not a property of the material, its a property of the entire object. Imagine having a small amount of a fairly tight spring. It takes a lot of effort to extend it even a centimeter or two. Now without changing the material, connect a few hundred of the springs together. Extending it a centimeter now will take ...


1

Law of inertia states an object in uniform motion continues in its state(i.e moving along straight line with uniform velocity) unless a force impressed thereon.. Mass is the measure of intertia of a body,i.e the massive the body greater the force required to change its state of motion... in the absence of ext.force the momentum (mv) of a object is ...



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