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The torque tau on a body about a given p...

The torque `tau` on a body about a given point is found to be equal to A x L, where A is constant vector and L is the angular momentum of the body that point. From this, it follows that

A

`(dL)/(dt)` is perpendicular to L at all instants of time.

B

the component of L in the direction of A does not change with time.

C

the magnitude of L does not change with time.

D

L does not change with time

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
(a,b,c) `tau = (vacdL)/(dt)`
Given that
`tau = rarrAxxrarrL rArr (rarrdL)/(dt) = rarrAxxrarrL`
From cross - product rule, `(rarrdL)/(dt)` is always perpendicular to the
plane containing `rarrA and rarrL.`
By the dot product definition
`rarrL. rarrL = L^2`
Differentiating with respect to time
`vecL. (dvecL)/(dt) + vecL. (dvecL)/(dt) = 2L (dL)/(dt) rArr 2vecL. (vecdL)/(dt) = 2L (dL)/(dt)`
Since, `(vecdL)/(dt) = 0 rArr (dL)/(dt) = 0`
`rArr L = constant`
Thus, the magnitude of L always remains constant.
As `vecA` is a constant vector and it is always perpendicular
to `vectau,`
Also, `vecL is perpendicular to vecA`
`:. vecL _|_ vecA :. vecL. vecA = 0`
Thus, it can be concluded that component of `vecL along vecA` is zero i.e., always constant.
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