# Tag Info

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See: http://en.wikipedia.org/wiki/Ehrenfest_paradox The Ehrenfest paradox asserts that for a spinning disc rotating at relativistic speed near the edge, the ratio of the diameter to the circumference is no longer pi. The proposed "resolutions" of this paradox have always seemed unconvincing to me. Like the pole and barn paradox, absolute rigidity or ...

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Let's suppose I have some system and I know $M$, the system's total mass, $\vec{r}_{cm}$, the system's center of mass position and $\vec{L}_{cm}$, the systems angular velocity in the frame where the center of mass is the origin. How do I find $\vec{L}'$, the angular momentum with respect to some other origin, say $\vec{r}_{cm} + \Delta \vec{r}$, which is ...

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Yes the skater does increase the angular momentum by doing work; pulling her arms in. You do work on a swing (sitting up and down) to increase your angular momentum likewise.

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It is the elimination of friction on the ground. The friction on the air is very small, as is the resistance of the rope to twisting. No matter how smooth the floor, the friction will be much higher than the resistance of the hanging weight. This is why air bearings were invented.

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So here are the equations of motion: You have the geometry so you know the mass moment of inertia about the center of mass to be $I_C=\frac{m}{12}\ell^2$ The applied force $f$ creates a torque about the center of mass equal to $\tau=\frac{\ell}{2}\,f$ The center of mass will accelerate by $a_C$ in the direction of $f$ with $$m a_C = f$$ The body will ...

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The string is essentially determining the radius of the uniform circular motion. So you just use the length of the string and the angle to find the radius of rotation (i.e. distance from particle to black dot). The relationship is just from right triangle trig $$\sin \theta = r/L$$ Then using Newtons laws with $a=v^2/r$ we get ...

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From the FBD, since there is no vertical acceleration we get, $mg = T \cos \theta \tag{1}$ where $T$ is the tension in the string From the circular motion of the bob, we get $\frac{mv^2}{L \sin \theta} = T \sin \theta \tag{2}$ Simplifying $(1)$ and $(2)$ $\tan \theta \sin \theta = \frac{v^2}{Lg}$ Infact, if you use angular velocity $\omega$ instead of ...

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Ok you have one degree of freedom here. Lets call it the angular position from vertical of the ball $\theta$. From that the center position of the ball is $\vec{r} = ((R-r) \sin \theta, R - (R-r) \cos \theta)$. The angular orientation of the ball is $\varphi = \frac{R}{r} \theta$ due to "gearing" between the two surfaces. The center of the ball tracks an arc ...

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Fictitious forces arise when you are working with non-inertial reference frames (e.g., rotating reference frames). Your problem statement includes the line What is the frequency of small oscillations? which means you are linearizing the system and no rotations are to be considered in the problem.

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Suppose the ramp wasn't there, then the trajectory of the object would the same as if it fell off a cliff: To get the equation of motion you simply note that the horizontal and vertical coordinates are given by (neglecting air resistance): $$x = ut$$ $$y = \tfrac{1}{2} g t^2$$ So you can get the trajectory by substituting for $t$ to get: $$y = ... 0 I clamped a small mousetrap to a heavy table and used a spring scale to measure the force when the hook was attached to the trap arm. Try to keep a 90 degree angle between the scale line of action and the arm of the mousetrap. The trap arm I measured was about 45 mm in length from the pivot point. I measured three forces, one to just lift the trap arm off ... 0 Ok.. I hope I understood it. Force of gravity is uniformly acting on the rolling body at every point of the body while it is rolling down. Suppose we push a rigid box on a surface then our force is acting on every point of it. The friction force will be opposite of its acceleration. Now for a rolling body it will depend on how we are moving the body. We can ... 0 First of all friction force is applied by the surface of the plane where the object is moving. when we walk on a surface we push the surface backwards as a result due to friction the surface push us which is the frictional force. It is in the direction we are walking. Now an object which is rolling with acceleration is rolling with increasing velocity. It ... 2 If the bumpers where vertical then the contact point would be at the center and since gravity is more than the bounce force it means there isn't going to be enough friction in change the rotation of the ball when the direction of the ball changes. If the contact point is further up, then the contact force is towards the center of the ball, and hence is ... 1 John Rennie's answer is great for DC motors but I thought I'd mention why AC induction motors reach a top speed. For AC induction motors, top speed is limited by the speed of the rotating magnetic field set up by the stator (aka, the synchronous speed of the motor). The rotor in an AC induction motor can only rotate as fast as rotating magnetic field and ... 4 Surpringingly the top speed is not necessarily anything to do with friction, though friction will of course have some effect. A motor acts as a generator, i.e. if you turn a motor it will generate a potential difference just like a generator, and this potential difference (usually called the back EMF) is proportional to the motor speed. So if you connect a ... 2 There are a couple of problems with this. One is that$$\tau_{net}=I\alpha$$is an equation that relates net torque to angular acceleration, but T shown in the figure is a force (tension), not a torque. So use the definition of torque to convert T into torque due to the tension (\tau_{T}). Also, the relationship between the linear acceleration a_{cm} ... 0 Took the following data: measured force on 0.045 m trap arm and computed torque (angle(rad), torque(n*m)) (0 , .1215) (1.57 , .2565) (3.14 , .378) best fit line to data torque = .0816 * angle + 0.1238 area under this line is energy stored in spring about 0.8 Joules 1 There are two parts to angular momentum that both contribute at the same time. In vector form (where × is the cross product)$$ \vec{H}_A = I_{cm} \vec{\omega} + \vec{r}_A \times m \vec{v}_{cm} $$For a horizontal rod rotating about end point A you have$$ \begin{aligned} \vec{\omega} & = (0,0,\Omega) & \vec{v}_{cm} &= \vec{\omega} \times ...

2

Good work and a good idea. d = L/2 would correspond to the moment of inertia for a point mass M at distance L/2. Momentum p = M $\omega$/2 L/2. What would you get if the mass of the rod was concentrated at the two end points, each 1/2 M ? One point zero, the other M/2 $\omega$ L. In other words d = 1. So the mass distribution along the rod plays a role. ...

0

Since you don't want a math description, but rather a physical intuition, I'll attempt to give one. The opposing force comes from the fact that when you try to rotate the gyroscope, you are attempting to redirect the path of the material in the spinning ring. Imagining the spinning ring as a discretized collection of point particles in motion, you are ...

3

If you assume that Your body is a uniform, thin rigid rod. One end of the rod is pivoted (aka your feet) during the fall. Then one simply recalls that the angular velocity $\omega$ of rotation of your body is related to the tangential velocity $v$ of a point a distance $r$ from the pivot by \begin{align} v = \omega r \end{align} Now, if you have height ...

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I think your analysis is good and correct. The statement you've quoted doesn't contradict what you've done, it probably suggests a different way to look at the problem. I think the statement simply states a method in mechanics- transferring a torque to another point (coordinate axes). In this case, from the centre of the wheel to the point of contact. ...

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This problem is an example of rolling without slipping. A very good explanation of this concept is given here. In this case, it implies that rolling without slipping occurs if $\tan \theta \leq 3 \mu$. The expression validates one's intuition too. Its easy to observe that a cylinder tends to roll without slipping when kept on a wedge with lesser slope.

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