Question

axis
43. A force F = ai +3j +6k is acting at a point
F = 21 -6° -12k. Find the value of a for which
angular momentum about origin is conserved.
(1) o
(2) 1
(3) -1
(4) 2
Two loons and are made from
ace,​

Answer 1

Answer:

  A = -1

Explanation:

The law of conservation of angular momentum states that the angular momentum of a body that is the product of its moment of inertia about the axis of rotation and its angular velocity about the same axis, cannot change unless an external torque acts on the system

In the given question F should be ,F= ai+3j +6k which will be acting at a point P 2î -6j-12k

For the conservation of angular momentum, w x I =0;

•°• i(2a+2) + j(6-6) +k(12-12)=0

•°• i(2a+2) = 0

   (2a+2) = 0

    2a = -2;

     A = -1

Answer 2

Answer:

(3) -1

Explanation:

1) We have,

$\vec{F}=ai+3j+6k\\ \\ \vec{r}=2i-6j-12k$

Angular momentum about origin is conserved when torque about origin is 0.

That is,

$\vec{r} \times \vec{F}=\vec{0}\\ \\=> \begin{vmatrix}i&j&k\\a&3&6\\2&-6&-12\end{vmatrix}=0\\ \\=>i*0+j(12+12a)+k(-6a-6)=0\\ \\=>12+12a=0,-6a-6=0\\ \\=>\boxed{a=-1}$

Hence, Value of a is -1.

Option (3) is correct answer.