
#1
Jan1108, 08:19 AM

P: 69

Can somebody explain to me clearly what is MOMENT OF INERTIA, what is TORQUE, what is ANGULAR MOMENTUM. If you ever comparing those with linear motion quantities, then please explain a little in detail what is the similarity between moment of inertia and mass, as I'm not getting it.
Also, I would also request you to answer this question.(it might help): What will happen if I am far out in space in a region without (or negligible) graviatational forces and try to spin a sphere. Would it spin for ever? (unless I do something) If it would, then each particle will undergo circular motion, which means that a constant centrifugal force is being applied towards the centre. If so, then who is applying it? Wouldn't that violate Newton's Second Law? Would it spin for sometime and then stop (refer Q.1) Any attempt of explanation is appreciated. 



#2
Jan1108, 08:41 AM

Mentor
P: 40,875





#3
Jan1208, 09:53 AM

P: 69

OK let me give it a try.
Since no external torque acts on it, it implies that it has no angular acceleration which further implies that it's angular speed would never decrease which even further implies that it should spin forever. The constant centrifugal force is applied by Electromagnetic attractions between the atoms (I call them bonds) which is always balanced by the opposite end. It actually spins forever due to Newton's Second law of rotation. Whenever we apply force on a spinning object it is (mysteriously) taken that all forces act on it's centre of mass (in this case the centre of the sphere). Don't ask me, Newton proved that we can indeed do this. Refer Q.1 If we apply a force which is not pointing to the axis of rotation, it produces a Torque and it would cause it to spin faster or slower. ***So, How was it**** (At least tell whether my concepts are correct) **any attempt of explanation is appreciated** 



#4
Jan1208, 10:04 AM

Mentor
P: 40,875

Some Questions regarding Rotational Mechanics 



#5
Jan1308, 04:07 AM

P: 69

Here's my second question:
Is angular displacement ([tex]\theta[/tex]) a vector quantity? If yes then why doesn't it follow laws of vectors. And what is the direction of this vector. if no then why is angular velocity ([tex]\omega[/tex]) a vector? 



#6
Jan1308, 06:03 AM

Mentor
P: 14,433

You can certainly ascribe three orthogonal components to describe any rotation. This is one interpretation of Euler's rotation theorem. However, angular displacement is not a vector quantity because the result of a sequence of angular displacements depends on the order in which the operations are performed. You can see that this is so by applying various rotation sequences to a book. The book will end up in one orientation if you first roll it by ninety degrees and then pitch it by ninety degrees. It will end up in a completely different orientation if you do the pitch first and roll second. To qualify as a vector, addition must be commutative (i.e. a+b=b+a), and angular displacements do not do that.
Suppose we describe a rotation as a roll,pitch,yaw sequence. Call this triple "a". Now do the same for another rotation, and call this sequence "b". As the rotations "a" and "b" become small, the difference in orientation between "a" followed "b" versus "b" followed by "a" becomes very, very small. In other words, in the limit of infinitesimally small rotations, "addition" of rotations is commutative. This is why we can view angular velocity as a vector quantity. Differentiating the noncommutative Euler sequences that describe a rotation paradoxically yields a commutative quantity. Technically, angular velocity is not a vector quantity. It does behave like a vector in the sense that angular velocities can be added like vectors. However, angular velocity does not behave like a vector upon reflection. Things that behave like vectors under proper transformations but not under improper transformations are called pseudovectors. 



#7
Jan1308, 10:47 AM

P: 69

Thanks for all your efforts.
Before I congratulate you for your wonderful efforts, I would like to add one final question. Let's suppose, we have a frictionless surface. On that a block with inclined plane of some mass say 'M' and length 'L' with inclination '[tex]\theta[/tex]' is kept. On it a sphere of mass 'm' and radius 'r' is kept andit rolls down the plane without slipping. The inclined plane has friction coefficient '[tex]\mu[/tex]'. There is going to be only one kind of friction and that is static friction. Am I correct? Will the block move (or accelerate) as the ball goes down? (Don't tell me to try that) ***Any attempts of explanation are appreciated*** 



#8
Jan1308, 12:07 PM

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#9
Jan1308, 12:22 PM

P: 733

first funny side: unfortunately in my 2nd, 3rd and 4th sem mechanics viva, i got the same teacher and the stupid person asked me all three times the same question, what do you mean by moment of inertia, angular momentum, their physical significance and analogy to mass and momentum. and everytime i explained so much, but still he would say, i lack my concepts.
anyways does the inclined block experience some reaction due to the sphere?? (in horizontal dir) 



#10
Jan1508, 10:21 AM

P: 69

Well according to me, the block should move behind, and the sphere shouldcome straight down if frictional force is less than gsin[tex]\theta[/tex].




#11
Jan1508, 06:33 PM

Mentor
P: 40,875

How can the sphere come straight down if the block moves to one side? (Consider conservation of momentum.)
Also: The sphere will roll down, not slide down. (Assuming the angle isn't too steep.) 



#12
Jan1608, 07:45 AM

P: 69

Then what will the motion of the sphere be like??




#13
Jan1608, 09:46 AM

P: 1

just read good refrence book like d.c.pandey mechanics2



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