### Re: Inertia in Linear and Rotational motion?

Posted:

**Thu Jul 30, 2015 12:39 pm**Even though you had already given me this link time in which you give me is appropriate

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Posted: **Thu Jul 30, 2015 12:39 pm**

Even though you had already given me this link time in which you give me is appropriate

Posted: **Sat Aug 01, 2015 8:00 am**

Let me assume a string and a ball connected with the spring. So if we rotate the system "ball-spring" around some center we get circular motion (no matter which plane (vertical, horizontal, tilted). Let's take a vertical plane so we have: T + mg = ma, hence T = m*(a - g). Then T = 0 when a = g (weightlessness of object). In this case if centripetal acceleration "a" goes to infinity it means that T increases (doesn't matter in which plane we rotate). If I don't mistake it looks like: plane of rotation has some relations of relativity with a vector (coming from center) of gravitational field? In other words, it depends on which angle is between the rotation plane and a gravity field radius I don't know how to express. I have not learnt gravity fields yet.

Posted: **Sat Aug 01, 2015 8:13 am**

To clarify my previous post I want to say the following. If we rotate the system in a vertical plane the equation is true: T + mg = ma (on the top of circular motion). Moreover, the state of weightlessness may be not only at the top. So if we rotate strongly enough we can get the state at other points around the top. However, if we rotates the system at a tilted plane with respect to the Earth this equation changes. Could someone explain how the equation changes? Clearly we additionally get x-components in this equation. But it's a circular motion so far. I suspect **x** and **y** components are not enough here, right?

Thank you.

Thank you.

Posted: **Sat Aug 01, 2015 12:50 pm**

For the rotation about an inclined axis, you'll need to specify components in all three dimensions.

Also, you need to be careful about two things:

- if the mass is on a spring, rather than an inextensible string, then its path will not be circular: the spring will be longer at the bottom.

- in the absence of other forces, it will not be uniform circular motion: it will travel fastest at the bottom.

So it will be a complicated problem.

Also, you need to be careful about two things:

- if the mass is on a spring, rather than an inextensible string, then its path will not be circular: the spring will be longer at the bottom.

- in the absence of other forces, it will not be uniform circular motion: it will travel fastest at the bottom.

So it will be a complicated problem.

Posted: **Mon Aug 03, 2015 1:32 pm**

Thanks a lot, Joe.

Posted: **Wed Feb 13, 2019 8:04 am**

It is not included in theMoment of Inertia is not applied in Linear motion at all, if I don't mistake. If it is so, why isn't Moment of Inertia applied in Linear motion?

I, which is the moment of inertia (also called rotational inertia), is the analogue of mass for rotational motion. So linear kinetic energy is 1/2 mv^2, rotational kinetic energy is 1/2 I omega^2.I think Inertia plays some role in Linear motion as well as in Rotational motion but Inertia is no any means taken into account in Linear motion formulas (Kinetic energy, momentum etc.). Why does it happen so?

http://www.animations.physics.unsw.edu. ... ation.html

Posted: **Wed Feb 13, 2019 8:06 am**

For many people, inertia and mass are synonymous. To avoid potential ambiguities, I usually avoid the term 'inertia'.What if define any mass by Inertia, not how it does now Inertia by mass? In other words, change a general method to model systems. Just thoughts.