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-1

It is because there aren't any forces acting on the mud keeping turning with the tire that it flies off. At whatever point the mud comes off, it will travel tangent to the tire at first and the follow a parabola due to earths gravity. It is most likely the more loose mud will come off first and at that point the tangential direction of the tire points ...


3

You were probably expected to note that the path of any point of the tire is a cycloid. At the point of contact with the ground, the tire is not moving. As it rises from the ground it is moving faster, with a speed that is $v(1+\sin \theta)$ where $v$ is the bike speed and $\theta$ is measured counterclockwise from horizontal. The centrifugal force is ...


-1

I'm not knowledgeable in some aspects of the question, but I will provide an answer unrelated to others. An object can rotate so fast that some representations of angular velocity cannot be valid. For example, when an object rotates more than 180-degrees or 360-degrees (pi or 2*pi radians) per unit time, the representation must be able to represent such ...


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If you're both going at a constant speed, lets say 35 MPH, then it would appear so. This is because the distance is shorter in the inner arc in which you are traveling compared to the outer arc in which the other vehicle is traveling.


0

I have read that the formula for angular velocity: $$\dot {\vec r}=\vec \omega \times\vec r \tag{1}$$ does not hold in some situations, but the book does not specify what situation so please could you produce a list of when this formula does not hold. That expression is only true in the case of circular motion. It fails whenever the radial component ...


1

If you have a single particle then you can indeed easily describe its motion using Newton's laws. However in rotating systems we are often dealing with continuous bodies not point particles. These are characterised by a moment of inertia, rather than just by a mass, and we have an equation analogous to Newton's second law: $$ T = I\dot{\omega} = ...


3

It would appear that the object is in circular motion at any given instant, which implies that the tension force provided by the string is always perpendicular to the velocity of the mass. Mostly true, but not quite. If the string is held steady, then the object is in circular motion. But by pulling the string harder than that, it moves off of ...


1

You are really asking about the reaction forces felt on a rod when you push on its end. For a rod, you can work with a quantity called the reduced mass (see for example this excellent answer for the derivation). As long as the rod is balanced on its end, the reduced mass tells you exactly how much greater the inertia of the cart appears to be: $$m_r = ...


0

Work done by the electric field is force times displacement along the direction of force. Center of mass doesn't move, so the only motion is rotation; horizontal spacing of charges goes from $a\cos\theta$ to $a$, so work done is $E q a(1-\cos\theta)$. Half of that work is done on the positively charged mass (which moves half that distance, i.e. ...


1

You will need to make use of the following: $$\cos^2\beta + \sin^2\beta\cos^2\gamma = \cos^2\beta + (1-\cos^2\beta)(1-\sin^2\gamma)$$ $$ = \cos^2\beta+1-\cos^2\beta-\sin^2\gamma + \cos^2\beta\sin^2\gamma$$ $$ = \cos^2\gamma + \cos^2\beta\sin^2\gamma$$ We therefore have $$\sqrt{v_\theta^2+v_\phi^2} = r\dot\gamma$$



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