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12

How can we detect Earth's spin? Apparent motion of Sun You will have observed that the sun reappears every 24 hours. There are two common explanations for this. One of them is that the earth rotates with a period of approximately 24 hours - this is the only explanation supported by the scientific evidence. The main alternative had a rather convoluted ...


5

There are several factors that may be taken in account, but the more important is the energy used deforming the tire. Suppose a deflated tire. As you move forward and the tire rotates, the part of the tire that is starting to touch the ground has to be deformed (since the tire is flat). You have to use an important amount of energy for that. Note that the ...


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As the object spins faster, you will need a higher drive frequency to continue to apply torque. At the size of a baseball, circumference is about 9.25 inches (let's say 25 cm). To rotate the surface at the speed of light would imply 1.2 GHz . At that frequency, the skin depth for steel is about 1.2 um, implying very high resistivity which would lead to high ...


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For a smaller object you can do a bit better "Dr Yoshihiki Arita, Dr Michael Mazilu and Professor Kishan Dholakia of the School of Physics and Astronomy at the University of St Andrews were able to levitate and spin a microscopic sphere, purely using laser light in a vacuum, briefly up to 600 million RPM before it broke apart." ...


3

Intuitively you can think of it this way: Moment of intertia measures how hard is to change the angular momentum of the object. Now, the angular momentum of a point single point of mass $i$ is: $$ \vec{L_i} = r_i \times m_i \vec{v_i} $$ To make this point rotate faster, you need to apply some torque. The mass part is easy, and one of the $r_i$ is just ...


2

There's another component to the solution: nonuniform friction. The blade, even if untaped, has different coatings than does the shaft. The butt end may or may not have a tape roll applied. All this means that, even if you throw the stick with zero spin applied, when it hits the ice it's almost guaranteed that there'll be a nonzero net torque due to the ...


2

Your claim is false (in $\mathbb R^3$). If $\vec{a}$ and $\vec{b}$ are fixed vectors (I assume they are not co-linear and satisfy $|\vec{a}|=|\vec{b}| \neq 0$) there are infinitely many rotations $R \in O(3)$ such that $R\vec{a}=\vec{b}$. One is the rotation $R$ of the angle between $\vec{a}$ and $\vec{b}$ performed around an axis orthogonal to the plane ...


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We generalize the result from the second answer you linked: $$\vec{L}=\sum_i \left(\vec{r}'+\vec{{r}_i}' \right)\times \vec{p_i}=\vec{r}' \times \sum_i \vec{p_i}+\sum_i \vec{{r}_i}'\times\vec{p_i}=\vec{r}' \times \sum_i \vec{p_i}+\vec L' $$ Now $\vec{r}'$ is some arbitrary vector, not just the vector to the center of mass. Thus when the system is moving ...


2

Mass is not a function of time, so the only thing inside the integral that needs differentiating is the position. Let me try to make this clearer. The integral over $m$ is really the integral over the volume: $$\int_m r \cdot dm = \int_V r \cdot\rho(r')~ dV$$ Here I deliberately distinguish between $r$, the vector from the origin of the coordinate ...


2

When two forces of equal magnitude, opposite in direction and parallel act on an object, the object will rotate without having translational speed. This condition is not enough. In rotational mechanics the position is also important. The point that makes this simple to overview is, that changes in translational motion is described by forces and e.g. ...


2

Perhaps the least convenient but the most direct is go to the Moon and observe the Earth. The (average) length of day as measured by timing stellar transit to stellar transit, sidereal day, differs by 4 seconds from the length of day defined by timing noon on one day to noon on the next day, solar day. A satellite launched East requires less energy to ...


1

In order for the spinning disc to set the stationary one in motion, friction forces need to act between the contacting surfaces: That requires a Normal force $F_N$ to act between them. Using the kinetic friction coefficient $\mu_k$ we can then state: $$F_F=\mu_k F_N$$ The friction force on the $m,r$ disc causes angular acceleration $\alpha=\frac{d ...


1

Yes you have to take the frictional force F into account so this exerts a negative impulse moment RFdt on the right hand disk with radius R as well as rFdt on the left hand disk with radius r. So the delta of the rotational moments have the ratio r/R. Assuming the left hand disk spins with rim velocity v0 before impact and after impact they have equal rim ...


1

The impulse is the change in total momentum of the body of mass. The momentum $P$ of a rigid body is the product of the velocity of the centre of mass $v$ and total mass $M$. A free rigid body's centre of mass translates at constant velocity, while the body itself performs a rotation about the centre of mass. Thus, after the impulse $J$ has been applied, ...


1

You should also note that static friction acts on the body when it is about to topple. Friction acts along the ground so it's torque about A is also 0. Torque due to F is clockwise and torque due to mg is counter-clockwise. Since torque= lever arm * force (or in vector form $\vec{\tau}=\vec{r}x\vec{F}$), torque due to mg is $mg\frac{a}{2}$. So net torque is ...


1

yes i think your assumption is correct based on the fact that in equilibrium the discs shouldn't slip on each other. but then you need to change your equation for angular momentum. use parallel axis theorem to write down the moment of inertia for the stationary disc to get the answer in the equation. since you have chosen your axis to be passing through the ...


1

When you drop a stationary disc onto a rotating one there must be a time when there is relative motion between the discs as you cannot have an infinite acceleration. If there is no friction then nothing much happens and the spinning disc carries on spinning and the other disc just sits still on top of it. To get an interaction between the discs you need ...


1

You can calculate angular velocity by taking origin ofyour frame on semicircle centre.unknowns as torque and friction force can be calcuted by balancing forces and writing torque equations knowing mass of beetle and semicircle.


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When there is curvature, tensile forces (in a ring, in a string, ...) will give rise to a net force as shown in the following sketch:


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It is called conservation laws, conservation of momentum and conservation of angular momentum. Because friction on the ice is very small, the geometry of the stick is a line with non uniform mass, there will be angular momentum and linear momentum that will be transferred to the ice at the points of contact. To only rotated there should be no linear ...


1

There seems to arrive much confusion on this topic. I've updated the answer here to give a clearer picture of what goes on. Consider a star instead of a ball rolling down the incline: For it to roll without slipping, whenever a leg is touching the ground it must stand still (it must not slip or slide). That means that during the time of contact for one ...


1

I think it's due to the fact that your transformation equations between $x_{\alpha ,i}$ and $q_{j}$ do explicitly contain the time. $$ x = Rsin(\phi )cos(\theta )=Rsin(\phi )cos(\omega t) $$ $$ y = Rsin(\phi )sin(\theta )=Rsin(\phi )sin(\omega t) $$ $$ z = Rcos(\phi ) $$ And, according to the theorem mentioned by Oiale, your kinetic energy does not present ...



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