# Search Results

Results tagged with Search options user 114696
114 results

A tag for questions about the mechanical interactions of rotating objects, including torque and angular momentum.

The normal and weight balance each other. You are making an assumption here. While the block is stationary the weight and normal reaction balance. But to make the COM start moving upwards there m …
answered Jun 11 '17 by sammy gerbil
Think of the bead on a large flat rotating disk. It rolls outward, and continues rolling outward if there is nothing to stop it. Why? Because of inertia. Or, in the rotating frame of the disk, because …
answered Aug 30 '16 by sammy gerbil
Updated Answer Apart from rolling friction (suggested by Noah) there are other reasons which might explain some of the discrepancy more simply. While rolling along the straight tube the ball bearin …
answered Dec 9 '16 by sammy gerbil
Throwing your arms back will throw your upper torso forwards, increasing the forward rotation. If this does "work" it is probably that by spreading your arms out (not throwing them back) you increase …
answered Feb 3 '18 by sammy gerbil
I think your problem is that you are trying to make an unnecessary (and unhelpful) distinction between circular motion and translational motion. When motion is divided between rotational and transla …
answered Jan 28 '17 by sammy gerbil
The sloshing about of the liquid water makes analysis of the motion of the bottle far too difficult. The centre of mass moves about, the moment of inertia (how easily the bottle can be rotated) change …
answered Sep 30 '16 by sammy gerbil
If the skew axis makes angles $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively then the moment of inertia about this axis for any rigid body is I=I_{xx}\cos^2\alpha+I_{yy}\cos^2\beta+I_{ …
answered Mar 9 '17 by sammy gerbil
You are confusing "rolling resistance" and friction. These are two different unrelated phenomena, so there is no reason why one should be less than the other. When the ball bearing rolls with constan …
answered May 13 '17 by sammy gerbil
Separate free body diagrams are required for the block $m_2$ and the pulley. That is why they are called free body diagrams : we isolate them from each other and consider only the forces acting direct …
answered Apr 30 '17 by sammy gerbil
You are almost correct. The linear acceleration of the end of the rod is $\alpha L$ perpendicular to the rod. The component of this acceleration in the direction of the string is $\alpha L\sin A$ wh …
answered May 4 '16 by sammy gerbil
Your terminology is confused. Rotational energy is a form of kinetic energy. I think you are distinguishing between rotational and translational KE. Also, torque is applied to the whole body, not one …
answered Feb 15 '17 by sammy gerbil
The CM can be used to calculate the change in the gravitational potential energy of the system in this example, but it does not always give the correct result. It works here because the gravitational …
answered Nov 3 '17 by sammy gerbil
The "trick" only works for elastic collisions, for which the relative velocities of approach and separation are equal and opposite, when measured along the common normal at the point of contact : $v_2 … answered Aug 3 '16 by sammy gerbil The Parallel Axis Theorem does not have a range. It applies regardless of how far an object is from the axis of rotation. However, as the distance$r\$ from the axis of rotation increases, treating th …
answered Nov 28 '17 by sammy gerbil
I assuming that you wish to find the distance from the thrower as a function of time. Suppose the thrower is at the origin, and the boomerang moves anti-clockwise in a circle of radius a with centre …
answered May 16 '16 by sammy gerbil

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