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Angular Momentum of asymmetric physical pendulum (Rigid Body)

Even if the body is constrained to rotate about the $z$-axis only, this does not mean that the angular momentum $\vec{L}$ must point along the $z$-axis. Rather, we always have $\vec{L} = \mathbf{I} \...
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Will the puck rotate without slipping in this situation?

The problem here is the assumption that $V_P=V_{CM}$ Your first thought that the motion of the centres of mass will be identical for the same total amplitude and direction of external force is correct,...
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Will the puck rotate without slipping in this situation?

There is no contradiction. $v_Q$ (the instantaneous horizontal peripherical speed) and $v_P$ (the horizontal speed of the string) are the same when $Q$ is at the top. But around every turn $v_Q$ ...
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Angular velocity of point on rigid body

As Mr. Hammen points out, the angular velocities are all the same. With an axis of rotation, the tangential velocity, v = rω, for each point.
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Number of Degrees of Freedom of a Rigid Body System - Proof

Definition. The dimension of the configuration space is called the number of degrees of freedom. Thus, if we find the dimension of the configuration space of a rigid body, we can deduce its degrees of ...
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Doubt in Equations of motion of rigid body, can anyone tell me how the second step occurred from first one

The answer @Robbie posted is the correct one. I wanted to add a helpful tweak to the cross product which might help you understand how to get to step 2 When you have a cross-product such as $\vec{\...
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Doubt in Equations of motion of rigid body, can anyone tell me how the second step occurred from first one

To me it looks like there might be a misprint. Easier to go straight from the first line. All it's doing is writing out the cross product explicitly: $$ (\omega \times J)_x = (\omega \times (I\cdot\...
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Is work done by torque due to friction in pure rolling?

Which answer is correct? The second answer is correct*. For some reason, friction tends to mentally twist people in knots. It is just an ordinary mechanical force and obeys all of the usual rules ...
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Conservation of energy in a rigid body

Yes, there are internal forces that hold the rod together. If there were no internal forces (imagine if the rod suddenly turned into sand), every piece of the rod would simply fall straight down.
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Is work done by torque due to friction in pure rolling?

Setting the second Newton's law of the situation, and using the definition of work: The net force for rolling without slip in an incline is $mgsin(\theta) - f_f$, where $\theta$ is the angle of the ...
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Is work done by torque due to friction in pure rolling?

Work is clearly done. The body speeds up and spins faster. First torque is the result of two parallel forces displaced sideways from each other. See Toppling of a cylinder on a block for more. One ...
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Is work done by torque due to friction in pure rolling?

Work depends on frame of reference. The reason is that, displacement also depends on frame of reference. We have three ways of analyzing work done by friction. If reference is ground, work done by ...
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Is work done by torque due to friction in pure rolling?

For a system of particles, the total kinetic energy (KE) is the KE of the center of mass plus the kinetic energy of the particles with respect to the CM. [Goldstein, Classical Mechanics] A rigid body ...
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Verifying directions of forces during pure rolling?

Your reasoning is completely correct. Here is an alternative line of reasoning. I usually find it simpler. Cause of motion is clockwise torque. So friction will apply anticlockwise torque to oppose ...
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Euler's Angles and Uniquely Defining the Orientation of a Rigid Body

Euler angles do not solve this problem 100%. There are cases where two or more sets of angles result in the same final orientation. We call these the degenerate cases, or gimbal lock, or singularities....
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How do the inertia tensor varies when a rigid body rotates in space?

This has been answered below, but consider the rotation matrix $\mathrm{R}$ whose columns represent the local $\hat{x}$, $\hat{y}$ and $\hat{z}$ axis in the world coordinates. $$ \mathrm{R} = \begin{...
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When two systems of forces acting on a rigid body are equivalent?

The equations of motion can be thought of given a specified motion, provided by the translation acceleration vector of the center of mass $\boldsymbol{a}_C$ and the rotation acceleration vector of the ...
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When two systems of forces acting on a rigid body are equivalent?

A theorem for a rigid body is: "Every system of forces is equivalent to a single force through an arbitrary point, plus a couple (either or both of which may be zero.)". A couple is defined ...
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Does an irregular rigid body can only rotate in three directions?

A general rigid body has 3 principal axis. Suppose $I_1<I_2<I_3$. If it rotates around $I_1$ or $I_3$ without external torques, the angular velocity doesn't change and is parallel to the angular ...
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Does an irregular rigid body can only rotate in three directions?

You state "an irregular rigid body has only 3 axes such that $ \vec{L}_{cm} $ and $ \vec{\omega} $ are parallel." I presume you mean the principal axes? In terms of the Cartesian principal ...
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Moment of Inertia of a solid hemisphere. What am I doing wrong?

Your symmetry argument is correct, this is because say we have full sphere and an axis passing through its center. Breaking it into 2 symmetrical hemispheres means each carries half of the total ...
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What is the angular momentum of a particle rotating around an axis in 3D?

The definition $\vec{L}_i = m \vec{r}_i \times \vec{v}_i$ is the correct one. Note that this vector will not lie along $\hat{n}$, and that's as it should be! Your derivation above makes the mistake ...
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3 votes

What is the angular momentum of a particle rotating around an axis in 3D?

they all take each particle's angular momentum to be ri×(mivi) instead of Ri×(mivi). Which is correct and why? I can understand your confusion since both terms do not seem to be the same. I think the ...
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Why is the sum of torques for each particle equal to the external torque?

I think I found a simpler way. First you prove $(\sum_i{\tau_i})\Delta\theta$ is the change in total kinetic energy for a small $\Delta\theta$. You know the change in a single particle's kinetic ...
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