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A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed \omega rad/s about the vertical. About the point of suspension :

 

  • Option 1)

    angular momentum is conserved.

  • Option 2)

    angular momentum changes in magnitude but not in direction.

  • Option 3)

    angular momentum changes in direction but not in magnitude.

  • Option 4)

     angular momentum changes both in direction and magnitude.

 

Answers (1)

best_answer

As we have learned

Angular momentum -

\vec{L}=\vec{r}\times \vec{p}

- wherein

\vec{L}  represent angular momentum of a moving particle about a point.

it can be calculated  as L=r_1\, P=r\, P_1

r_1 = Length of perpendicular on line of motion

P_1 = component of momentum along perpendicualar to r

 

 

 Angular momentum = mvr is constant as distance of bob from point of suspensions and its speed both are constant.

Direction of angular momentum changes continually as shown in figure 

 


Option 1)

angular momentum is conserved.

Option 2)

angular momentum changes in magnitude but not in direction.

Option 3)

angular momentum changes in direction but not in magnitude.

Option 4)

 angular momentum changes both in direction and magnitude.

Posted by

SudhirSol

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