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9

I know that kinetic energy is conserved so $\ \Delta KE=\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=0$ and that $\ I_{end}=\frac{1}{3}Ml^2$ which in this case is $\ \frac{1}{3}*2kg*1m^2=\frac{2}{3}$ You are looking for $\omega (= y)$ from conservation of $KE = (1*3^2/2)$you know that $\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=0\rightarrow 1*3^2 = y^2 ...


6

Now, after the hammer is released, the thrower still has her same angular momentum (also 62.8), but the hammer no longer seems to have any. A body does not have angular momentum wrt to a point C only when it is circling around it, you know that planets have elliptical orbits and do have L If a body H has linear momentum p it has also and angular ...


6

The rotational energy of a body is given by: $$ E = \tfrac{1}{2}I\omega^2 $$ where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a uniform sphere the moment of inertia is related to the mass of the sphere, $m$, and the radius of the sphere, $r$, by: $$ I = \frac{2}{5}mr^2 $$ You already have the mass, and you can Google for the ...


5

No, in general they do not. You can work this out from the geometry of Ackermann steering, discussed on my website article "Parallel Parking a Car". In summary: look at the defining geometry for Ackermann steering, which I have sketched below: Ackermann steering is defined by the intersection of the central unit normals to (axes of rotational symmetry ...


4

What I cannot understand is, why acceleration, a=lθ¨ and not lθ¨/2? The equation you wrote doesn't mention anything about the linear acceleration. Is the center of mass located at its top and not the center? Or is there something else I am missing? The center of mass of the pencil is in the middle, not the top. There is likely something else ...


3

"Equation that is all over the internet"... You started at http://thatsmaths.com/2014/06/26/balancing-a-pencil/ and from there, you linked to http://arxiv.org/pdf/1406.1125v1.pdf which was the source for the former. In the third paragraph of that paper, it states We model the pencil as an inverted simple pendulum with a bob of mass m at one end of ...


3

Cover it in mirrors that are highly reflective on one side and painted black on the other. Position the mirrors so that the "faces" are perpendicular to the surface. A sketch is below (I have only shown three mirrors, the idea is that you would cover the planet with them, but they will be most effectively placed close to the equator). The plan is that each ...


2

First, let's review the basic ideas of simple harmonic motion (I'm assuming an early university level). Starting with Newton's equation: $$F=ma$$ and using Hooke's law $$ma=-kx$$ then recognizing that acceleration is the second derivative of position x $$mx''= -kx$$ We know that simple harmonic motion is sinusoidal, so we substitute $x=\sin(\omega t)$ ...


2

Now, after the hammer is released, the thrower still has her same angular momentum (and has to slow herself down), but the hammer no longer seems to have any. Even though the hammer isn't rotating around the axis, it still has the same angular momentum it had at release with respect to the original axis. So the formula $$L = mvd$$ is correct both for ...


2

Yes. The solution is: $$ \bf{r} = \dfrac{\left( \sum {\bf F}_i\right) \times \left( \sum ({\bf r}_i \times {\bf F}_i) \right)} {\| \sum {\bf F}_i \|^2} =\dfrac{{\bf F} \times {\bf \tau}}{{\bf F}\cdot{\bf F}}$$ Then you can show that $$ {\bf r}\times \left(\sum {\bf F}_i \right)= \sum ({\bf r}_i \times {\bf F}_i) = {\bf }\tau$$ Use ${\bf F} =\sum {\bf ...


2

The rotational kinetic energy of a (uniform) solid sphere rotating about an axis passing through the center of mass is given by $\frac{1}{2}I\omega^{2}$, where $I=\frac{2}{5}MR^{2}$. So $K=\frac{1}{5}MR^{2}\omega^{2}$. Using $M=6\times10^{24}\,\mbox{kg}$, $R=6400\,\mbox{km}$, and $\omega=\frac{2\pi}{T}$, with $T=24\,\mbox{hrs}$, we get ...


2

You forgot to multiply $T \sin{\theta}$ by the distance from the wall to the end of the bar in the torque balance. When you do that, you get an extra factor of 4 in the first term for the expression for x, $x = \frac{8\sin{\theta}}{\mu_s \cos{\theta} + \sin{\theta}} - 2$, which is positive. (PS: I didn't check your math, I just added the factor of 4, so I ...


2

The answer depends on the direction of the axis of rotation. If the axis is normal to the plane, then you have the same amount of material the same distance from the axis of rotation as before - and thus the moment of inertia about that axis would be unchanged. However, if the axis of rotation you consider is in the plane of the paper, the answer will ...


1

Use the hint in your second bullet point. The ball has angular momentum about the pivot point before it strikes the stick.


1

Your conservation of kinetic energy equation should help you solve the for the stick's initial angular velocity. Think of it this way: the tennis ball has initial momentum since it is moving, right? And the stick is not moving, so it has no momentum. At the end of the collision, the tennis ball stops completely, so it has no momentum, but the stick is ...


1

Angular momentum is conserved in this example! As you already stated, the angular momentum of the thrower doesn't change after the hammer is released. Consider the hammer being in rotation around the origin of our coordinate system for $t < 0$: $$ \vec{r}(t) = r_0 \ \ (cos(\omega t), sin(\omega t), 0)^T $$. Its momentum is therefore given by: $$ ...


1

The fastest way is to compare kinetic energies in the two cases: \begin{align*} KE &= \tfrac{1}{2}I_{\text{cm}}\omega^2_{\text{cm}} + \tfrac{1}{2}M(R\omega)^2_{\text{cm}} \\ KE &=\tfrac{1}{2}I_{\text{inst}}\omega_{\text{inst}}^2 = \tfrac{1}{2} (I_{\text{cm}} + MR^2)\omega^2_{\text{inst}} \end{align*} So $\omega_{\text{inst}}=\omega_{\text{cm}}$. The ...


1

There absolutely is a period of acceleration. Speed never changes instantly, even if it changes too quickly for you to sense with your eyes and ears, as a direct consequence of Newton's laws. Probably it accelerates over a 1/10 of a second or so, if I had to guess.


1

I don't think that equation is right. $F_{max} = \mu mg$, so $a_{max} = \mu g$. Where are you getting velocity from? The spring doesn't move at a constant velocity, does it? You need to use the spring's maximum torque and work out how to weaken it so the final acceleration is sufficiently low. Why are you trying to make it as fast as possible? If you're ...


1

Let us see a similar example: two people on skates going with some velocity towards each other both a bit on left off their common center, and in the moment of the closest approach, they just catch each other by right arms and they start to rotate. In fact they have (as one system) the same angular momentum all the time. When you have a projectile that ...


1

You are really close. At this point you have an expression for $\tau$ which you need to minimize with respect to $x$ - you just didn't take the last step, which is writing $I$ as a function of $x$ in that expression: $$\tau = 2\pi \sqrt{\frac{\ell^2/12 + x^2}{gx}}$$ A minimum / maximum will occur when $\frac{d\tau}{dx}=0$. To keep your life simple it's ...



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