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-1

it's not clear to me why friction wouldn't impart a torque counter to the torque due to gravity.. why isn't Net Torque = wrsin(theta) - (F x R) where F is friction force?


3

There is a difference - but not exactly why you think. There are prevailing winds around the earth - these used to be called the "Trade Winds" because traders, knowing the direction of the wind, knew how best to navigate the globe. Basically, on the equator (in the tropics) they flow from east to west, and at higher latitudes they flow from west to east: ...


1

Start digging at the equator and move all the dirt to the polar regions. This will decrease the moment of inertia of the planet about its spinning axis. Due to the conservation of angular momentum this will result in an increase in angular velocity, akin to a figure skater who retracts her arms while spinning.


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 ...


0

I believe your second interpretation is spot on. Whenever anything rolls without slipping, it means that the point of contact of the ball with the frictional surface is instantaneously stationary. Therefore, the frictional force applied to this point does not do any work, and so there is no frictional dissipation. Also, because the ball is still rolling ...


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 $x=\sin(\omega t)$ ...


0

Note: I see there's been quite a bit of activity while I was typing this up. I'll leave the answer here for now. it doesn't seem immediately intuitive how the problem works Assume the spring provides a linear, tangential restoring force for small angular displacement (rotation) of the disk. The intuition here is that the problem will 'look' like an ...


0

The increase in rpm depends upon angular acceleration which is how quickly it's rpm increases (in this case) or decreases. When u increased the rpm, the change appeared instantaneous because of very small time required to increase the rpm. The less the time required, the greater tbe acceleration. So it is a period of acceleration.


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

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 ...


0

After the force $\vec{F}$ is applied on the bar, there could be two reasons for the acceleration of point $A$: Linear acceleration of the bar, Angular acceleration of the bar at point $A$. We can think of the bar as a point particle with all its mass collected at the center of mass (the point $C$ in this case) so that the first contribution I mentioned ...


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 ...


0

If friction and air resistance are negligible(if you want them in the calculations,just say it because now i am just assuming that this is the simplest problem) and the spheres start rolling from the same point,then the law f conservation if energy states that Mgh=0.5MV^2.You have M in both sides,so you can see that the final velocity is not depended on the ...


0

If the vehicle has 4 wheels, and only 2 have traction then the formula is $$ \begin{cases} a_{max} = \gamma \mu g & \mbox{RWD}\\ a_{max} = (1-\gamma) \mu g & \mbox{FWD} \end{cases}$$ where $\mu$ is the traction coefficient, $g$ is gravity and $\gamma$ is the %weight on the back wheels. So if the weight distribution is 60%/40% front to back, then ...


0

You cant avoid skidding by an equation. Would You glue a paper with that formula to Your car? You avoid skidding by using the appropriate tires for the floor. Look here for a solution for a car going as far as possible: en.wikipedia.org/wiki/Mousetrap_car


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 ...


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 ...


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: $$ ...


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 ...


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

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


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

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 ...


0

You may want to put the moment of inertia $$I(x) = a + b x ^ 2$$ into the formula for $\tau$, then differentiate with respect to $x$ and find out the $x$, for which $$\frac{d \tau}{dx} = 0$$


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

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 ...


-1

When the hammer thrower swings the hammer, according to Steiner's Theorem, the moment of inertia of the system is combined from a mass point at a distance from the center of rotation and the body rotating around its center of mass $$I = I_c + m \times r^2$$ so our complete moment of inertia is combined from 4 parts. $$I_{\text{complete}} = (I_{c_1} + m_1 ...


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 ...


0

This is a nice example which shows understanding does not come automatically after completing a calculation. But calculation still serves the (perhaps the most) important guide. Nobody in the above has mentioned the discussions given in \ittext{Landau & Lifshitz, Mechanics (BH, 3rd ed.), page 112}. I think these discussions have already elucidated the ...


0

Let me introduce the notation $$\sum F_{i,x} = F_x, \ \ \ \sum F_{i,y} = F_y, \ \ \ \sum F_{i,z} = F_z, \tag{i}$$ Since the determinant is zero, there may be indeed, no solution of the system. But if the system of equations has a solution, recall that the body doesn't rotate around a point, but around an axis. So, your $\vec r$ is bound to be on an ...


0

If you take the perspective that "work is the increase in total mechanical energy of a system", then there really no work being done. The potential energy is just being converted to kinetic energy (of some sort). It's not really correct to say that " total kinetic energy (translational and rotational) is mgh". It would be more correct to say that any ...


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 ...


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 ...


0

When a ball rolls down a ramp, it will accelerate more slowly than a sliding object (without friction). This suggests that you could treat the motion as that of a sliding object with a larger apparent inertial mass. I believe you will be able to solve your problem if you take this hint and apply it to your situation. You might find some inspiration in this ...


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 ...


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 ...


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 ...



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