# Tag Info

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You need to calculate the moment of inertia of the two systems separately. For rotation about own axis, you will have one inertia. For rotation about Sun you will have one moment of inertia. Moment of inertia is always defined with respect to axis of rotation. But yes, the object will have a energy or angular momentum which is a summation of energy and ...

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This is done using the parallel axis theorem $$I = I_{cm} + mr^2$$ where $I_{cm}$ is the moment of inertia of the disc and $I$ is the moment of inertia with respect to the athlete, $m$ is a mass of the disc and $r$ is the perpendicular distance between the athlete’s axis and the axis of the disk. So if you can calculate the moment of inertia of the disc and ...

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No, it won't be the same. Always remember to specify the axis with respect to which you are evaluating the moment of inertia. Usually we calculate the moment of inertia with respect to the center of mass which is the smallest possible of your rigid body. So it would be more correct to write $I_0$ for the moment of inertia or $I_{CM}$. There is a theorem that ...

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No, it wouldn't not be same as Moment of Inertia will change . $\omega$ is same for every particle in a rigid body . Hence , $L_{axis}=I_{axis}\omega$ for a rigid body in pure rotation.

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If you look carefully at the Integral you want to perform, you notice there is no angle dependence. So you can move pieces of you sphere around, as long as r of each individual element stays the same. By using rotations, you can rotate each of the 'quarter' elements of the sphere to exactly match your object. Long story short this means: I(Sphere)=4I(your ...

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Yes. If you know the shape and density you can calculate the inertia about a fixed axis or about a point. I have edited this answer to respond to comments received. To evaluate rotation about a fixed axis (or more generally, general rotation in a plane), you use the moment of inertia about that axis; this moment of inertia is a scalar. Let the axis of ...

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Of course you can! You "simply" have to use the definition of moment of inertia. The difficult is doing the right integration. Remember to specificate always respect which axis are you calculating the moment of inertia to. In the picture it's shown. Also, should be specified if the rigid body has an uniform density. Let assume it is. In this case ...

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Yes, use symmetry arguments for this particular case(think of how this setup is different from a sphere whose $I_{CM}$ is known). Otherwise in general case integratation is always an option, it may be complicated or you may need to solve the integral numerically but it will definetely give you the answer.

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Yes. It is always possible to calculate the moment of inertia of any arbitrary distribution of mass by doing a volume integral, summing up the moment of inertia for every infinitesimal mass $dM=\rho\,dV$: $$I=\int dM\,r^2 =\int \rho\,dV\,r^2.$$ Here $r$ is the distance of each infinitesimal mass from the axis of rotation. Sometimes this integral cannot be ...

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Please check this out http://stemandmusic.in/physics/moiNotes.php Hope this will help

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Remember that saying "moment of inertia" is actually a short for "moment of inertia of object about some axis." Therefore, the moment of inertia does indeed depend on the axis of rotation. I leave the computations up to you, but in some cases it may be that the moment of inertia will not change, so the period will not change.

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First let's take the situation when the object is moving without rotation, and by some internal mechanism, one of the weights moves. For an inertial observer, by conservation of momentum, the other weight (together with the rod that connects both weights) must also move, keeping the COM in the same previous straight line. If the same happens for a rotating ...

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Consider the following system, moving on a flat and frictionless surface. A uniform bar has two masses $1$ and $2$. $1$ is attached to the bar (rigidly), $2$ can move freely and without friction along the bar. We now engineer it so that at $t=0$ the CoM translates with velocity $v$ and the bar plus masses system rotates about the CoM with angular velocity $\... 0 To see that your thesis is true , let calculate the collision case of this example . The two pendulums are start with initial angular velocity and collide. I start with writing the equation of motion shortly after the collision , you have to deals just with the constraint force. $$I_1\,\ddot \varphi_1=L\,F_c$$ $$I_2\,\ddot \varphi_2=-L\,F_c$$ where$~I_1\,...

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Think about the $M d^2$ term as follows: The velocity of a mass at the end of a rotating massless rod is proportional to $d$ $$v = \omega d$$ Thus the translational momentum of said mass is proportional to $M d$ $$p = M v = M (\omega d)$$ The angular momentum of the mass ("torque" of momentum) is proportional to $d p$ $$L = d M (\omega d) = (M d^2)... 1 Why is the moment of inertia of a point mass defined as mr^2? This is a good question. If we wanted to understand this quantity I = mr^2 which has some definition, the first thing we could do is think about what the definition means. If taken literally, saying that I = mr^2 = r (mr) is the 'moment of inertia' of a particle actually implies (see below) ... 0 A point mass is orbiting about the origin with rotational velocity \vec{\omega}. If the particle location is at \vec{r}, then its speed is \vec{v} = \vec{\omega}\times\vec{r}, where \times is the vector cross product. The momentum of the particle is \vec{p} = m \vec{v} = m (\vec{\omega} \times \vec{r}) The moment of momentum (angular momentum) is \... 0 After thinking about it some more, I have realized that my mistake was not varying the relative masses through their entire range. Suppose that the payload is large compared to the ring (like, orders of magnitude more massive). In this scenario, the ring will essentially spin around the payload (because the barycenter will likely be inside of it), and when ... 0 To address this question let me first provide some necessary background. This image offers a representation both of Newton's first law, and an area law. An object, released to move freely, will move along the equally spaced points ABCDE in equal intervals of time. Additionally you have the following option: Taking an arbitrary point S that moves in inertial ... 9 True the distance from the axis has decreased but notice that we are taking the square of the square of the distances and hence the signs go away. That is, even if the average change of a collection of quantities is zero, the sum of squared quantities would not be unless all the quantities under consideration is zero. There is actually parallels between ... 6 Consider for simplicity just two "particles" of mass m, each on a massless rod of length, say 2L. The moment of inertia of this system wrt the COM is \begin{equation} I = mL^2+mL^2=2mL^2. \end{equation} Now move the axis by x. Then the moment of inertia becomes \begin{equation} I = m(L+x)^2+m(L-x)^2=2mL^2+2mx^2. \end{equation} The crucial thing ... 1 If you want to find the moment of inertia of the three-quarters disc about the centre of the original disc then you can just subtract the moment of inertia of the quarter disc about this same point. So you are left with three quarters of the moment of inertia of the original disc. If you want to find the moment of inertia of the three-quarters disc about its ... 3 Assuming that angular momentum was conserved throughout this expansion process then$$I_i \omega_i = I_f \omega_f$$where I_i, I_f are the initial and final moments of inertia, and  \omega_i,  \omega_f are the initial and final angular velocities. Now since$$\omega = \frac{ 2 \pi}{T}$$where T is the period for one revolution, you can write$$2 \...

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