If - - L- - 0 for a rigid body - where - - resultant torque L- - angular momentum about a point and both are non - zero- Then

The correct option is D |L⃗| = may increase ...

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Byju's Answer

Standard XII

Physics

Newton's Second Law: Rotatory Version

If τ⃗ × L⃗ ...

Question

If $\to τ \times \to L = 0$ for a rigid body , where $\to τ$ = resultant torque & $\to L$ = angular momentum about a point and both are non - zero. Then

\to L

= constant

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| \to L |

= constant

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| \to L |

= will increase

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| \to L |

= may increase

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Solution

The correct option is D $| \to L |$ = may increase

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Similar questions

Q. If

\to τ \times \to L = 0

for a rigid body, where

\to τ =

resultant torque &

\to L =

angular momentum about a point and both are non - zero. Then

Q. The torque

\to τ

on a body about a given point is found to be

\to A \times \to L

where

\to A

is a constant vector and

\to L

is angular momentum of the body about that point. From this it follows that

Q. The torque

\to τ

on a body about a given point is found to be equal to

\to A \times \to L

where

\to A

is a constant vector and

\to L

is the angular momentum of the body about that point. From this it follows that :

Q. The torque

τ

on a body about a given point is found to be equal to

\to A \times \to L

, where

\to A

is a constant vector and

\to L

is the angular momentum of the body about the point. From this, it follows that

τ = \frac{d \to L}{d t}

(where

L

is the angular momentum and

t

is time).
a)

\to A

is perpendicular to

\to L

at all instants of time.
b) The component of

\to L

in the direction of

\to A

does not change with time.
c) The magnitude of

\to L

does not change with time.
d)

\to L

does not change with time.

Q. The angular momentum of a particle relative to a certain O varies

\to L = \to a + \to b t^{2}

where

\to a

and

\to b

are constant vectors with

\to a

perpendicular to

\to b

. The torque

\to t

relative to the point O acting on the particle when angle between

\to τ

and

\to L

45^{o}

is:

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If →τ×→L=0 for a rigid body , where →τ = resultant torque & →L = angular momentum about a point and both are non - zero. Then

The correct option is D |→L| = may increase

If $\to τ \times \to L = 0$ for a rigid body , where $\to τ$ = resultant torque & $\to L$ = angular momentum about a point and both are non - zero. Then

The correct option is D $| \to L |$ = may increase