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A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad/s, the magnitude of its angular momentum about a point on the ground right under the center of the circle is
A. \[14.4g{{m}^{2}}{{s}^{-1}}\]
B. \[11.52kg{{m}^{2}}{{s}^{-1}}\]
C. \[20.16kg{{m}^{2}}{{s}^{-1}}\]
D. \[8.64kg{{m}^{2}}{{s}^{-1}}\]

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Last updated date: 29th Mar 2024
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MVSAT 2024
Answer
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Hint: An object in motion will in general stay in motion at a steady velocity except if followed up on by a force. That is on the grounds that momentum is saved.

Complete step-by-step answer:
A moving object has momentum, so it needs to continue moving at a similar velocity all together for its momentum to remain the equivalent. The main route for the object to quit moving is for it to move its momentum to another object. (That is the thing that force is: it's the exchange of momentum.)
The correct answer is A.
\[\overset{\to }{\mathop{r}}\,=x\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\,+z\overset{\wedge }{\mathop{k}}\,=r\cos \theta \overset{\wedge }{\mathop{i}}\,+r\sin \theta \overset{\wedge }{\mathop{j}}\,+z\overset{\wedge }{\mathop{k}}\,=0.1(\cos \theta \overset{\wedge }{\mathop{i}}\,+\sin \theta \overset{\wedge }{\mathop{j}}\,)+0.8\overset{\wedge }{\mathop{k}}\,\]

where θ is the angle the radius makes with the x-axis as the particle moves in the circle.

\[\overset{\to }{\mathop{p}}\,=-m(r\omega )(\sin \theta \overset{\wedge }{\mathop{j}}\,+\cos \theta \overset{\wedge }{\mathop{i}}\,)\]
\[\overset{\to }{\mathop{L}}\,=\overset{\to }{\mathop{r}}\,\times \overset{\to }{\mathop{p}}\,=(0.6(\cos \theta \overset{\wedge }{\mathop{i}}\,+\sin \theta \overset{\wedge }{\mathop{j}}\,)+0.8\overset{\wedge }{\mathop{j}}\,)\times (-mv(\sin \theta \overset{\wedge }{\mathop{j}}\,+\cos \theta \overset{\wedge }{\mathop{i}}\,))=2\times (0.6\omega )\]

\[2\times 0.6\times 12\]

\[=14.4kg{{m}^{2}}/s\]

Similarly, when you set an inflexible body in pivot, it will in general keep turning at a similar speed, around a similar axis - except if followed up on by a force.

We state that turning bodies have angular momentum. Angular momentum is moderated, and the main path for an object to quit pivoting is if its angular momentum is moved to another object. (That is the thing that force is: the exchange of angular momentum.)

Presently, in view of genuine experience, we realize that revolution isn't generally "saved". For instance, if you hold a coin with one finger so it remains on its side, and afterward you flick it with a finger from your other hand, the coin will begin turning. It didn't get its angular momentum from something different that was turning first. It was the flick that set it into turn.

When an object moves past an axis, it has angular momentum around that axis. In any case, when the object travels through an axis, or corresponding to the axis, it doesn't have angular momentum around that axis. You need to flick the edge of the coin, with the goal that your finger doesn't go through the coin's width. That way your finger will have angular momentum around the coin's distance across, and the coin will pivot about its own width as the axis.

If you flick the focal point of the coin, the coin will at present move, as direct momentum would be moved from your finger. In any case, the coin won't turn, as your finger won't have any angular momentum to move to it.

Note: Angular momentum relies upon the axis you decide for estimating it. This makes angular momentum more confounded than straight momentum.
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