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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

The coefficient of static friction and the normal force together allow you to calculate the maximum force that friction can apply, not the actual force. A block sitting on a shelf with no external horizontal forces will have a friction force of zero. If you apply a force from the side that is less than the maximum friction force, then the actual friction ...


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 ...


3

Your first quote is correct for an idealised model. There is no rolling friction then. Both wheel and surface are considered completely rigid. Ideal model - no rolling friction Non-ideal/more realistic model - rolling friction comes into the picture These pictures are from this link that gives a very good graphic view on this. Going away from an ideal ...


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

Friction is essentially the name people assign to the electromagnetic interactions on the everyday scales. Your question then comes down to asking what the universe will be if no electromagnetic interaction were present, which is the same as asking what the universe will be if we replaced the interactions we currently have with something else. The answer is ...


1

'Rolling friction' isn't actually a frictional force in the same sense as sliding friction. The term 'rolling friction' is a little misleading, and is better named 'rolling resistance'. When something rolls, e.g. a wheel on an axis, a lot of forces go into resisting the rolling. These forces can include the friction of the bearings, the momentum of the tire, ...


1

So one useful way to view friction is to imagine that the surfaces of the materials we're talking about are loaded with a bunch of springs which, after they hit a certain displacement, "break", stealing a constant vibration energy of $\epsilon = k \lambda^2/2$ over some length scale $\lambda$ that they need to break. The exact functional form of this energy ...


1

The sign of friction depend on our choice of axis. If I choose right side to be positive and I push an object to the right then friction will act to the left and so the friction will be negative. But the magnitude of friction is always positive. It is positive by definition. The magnitude of any force (or any vector) is positive. In your question you ...


1

You are correct. Kartik's answer is also correct, but he just didn't directly answer your question (well, to be fair, it presupposes a complete rewriting of the laws of physics). A positive coefficient of friction opposes the direction of applied force, reducing or preventing movement. With a stationary object, the force is called static friction, and for a ...


1

Friction occurs whenever energy is dissipated as movement happens. Suppose we move a distance $x$, and some mechanism dissipates an energy $E$ as we move (we won't worry about the mechanism of the dissipation for now). That means to maintain a constant speed we have to put in an amount of work $W = E$. But since we know that work = force $\times$ distance, ...


1

You are right. In an optimal system, there isn't any friction. But in real life there is, because there aren't any balls or surfaces that are so perfect that they touch just in one point. Also the surface of the ball is never perfectly flat, so there will always be friction between the surface of the ball and the surrounding air. So for calculating the ...


1

Do a free body diagram and you will find for a horizontal plane that $$ F - \mu m g = m a $$ $$ (100) - \mu (50) (9.80665) = (50) (0.1) $$ $$\boxed{ \mu = \frac{(100)-(0.1)(50)}{(50)(9.80665)} = 0.1937\ldots }$$


1

When you push the block the block 'pushes' you with the same force and you both gain equal and opposite direction momentums. Both block, and you have now some momentum and hence kinetic energy. The work done on both you and the block is: $$ W=\int F \,dx$$ where $F$ is a force applied. It must be equal to the total kinetic energy of you and the block (if ...


1

I would treat the car and the Earth together as a system. No internal forces can change the momentum of that system. However, the brakes can change the relative motion of two parts of that system - stopping the motion of the car with respect to the Earth.


1

The volumetric flow rate has to stay the same, because the same amount of water flows out of the tube as flows into it. The only way for the volumetric flow rate to change along the tube would be if there was a leak. Friction (i.e. viscous drag) can't change the volumetric flow rate, all it can do is change the pressure gradient along the tube. So you are ...


1

dmckee's comment is correct, this problem is hard (impossible?) to solve with pen and paper. But it's not too hard to set up, which hopefully will answer your question. There will advanced calculus by high school standards (including differential equations), no way around that, but I'll try to explain the equations at a high school level, I hope... I'll ...


1

It is an implicit function which you have to solve numerically. Typically you would use "fixed point iteration". You can do this in Excel with some user defined functions where for a given geometry you start with $f=1$ and then use the above $\frac{1}{\sqrt{f}} = L(f)$ a few times until in converges to a value $$f \rightarrow \frac{1}{L^2(f)} $$ In my ...



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