A force $$F = \alpha \hat i + 3\hat j + 6\hat k$$ is acting at a point $$r = 2\hat i - 6\hat j - 12\hat k.$$ The value of $$\alpha $$ for which angular momentum about origin is conserved is
A.
$$-1$$
B.
$$2$$
C.
zero
D.
$$1$$
Answer :
$$-1$$
Solution :
When the resultant external torque acting on a system is zero, the total angular momentum of a system remains constant. This is the principle of the conservation of angular momentum.
Given, force $$F = \alpha \hat i + 3\hat j + 6\hat k$$ is acting at a point $$r = 2\hat i - 6\hat j - 12\hat k$$
As, angular momentum about origin is conserved.
i.e. $$\tau = {\text{constant}}$$
$$ \Rightarrow {\text{Torque,}}\,\tau = 0 \Rightarrow r \times F = 0$$
\[\left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
2&{ - 6}&{ - 12}\\
\alpha &3&6
\end{array}} \right| = 0\]
$$\eqalign{
& \Rightarrow \left( { - 36 + 36} \right)\hat i - \left( {12 + 12\alpha } \right)\hat j + \left( {6 + 6\alpha } \right)\hat k = 0 \cr
& \Rightarrow 0\hat i - 12\left( {1 + \alpha } \right)\hat i + 6\left( {1 + \alpha } \right)\hat k = 0 \cr
& \Rightarrow 6\left( {1 + \alpha } \right) = 0 \cr
& \Rightarrow \alpha = - 1 \cr} $$
So, value of $$\alpha $$ for angular momentum about origin is conserved, $$\alpha = - 1$$
Releted MCQ Question on Basic Physics >> Momentum
Releted Question 1
Two particles of masses $${m_1}$$ and $${m_2}$$ in projectile motion have velocities $${{\vec v}_1}$$ and $${{\vec v}_2}$$ respectively at time $$t = 0.$$ They collide at time $${t_0.}$$ Their velocities become $${{\vec v}_1}'$$ and $${{\vec v}_2}'$$ at time $$2{t_0}$$ while still moving in the air. The value of $$\left| {\left( {{m_1}{{\vec v}_1}' + {m_2}{{\vec v}_2}'} \right) - \left( {{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2}} \right)} \right|$$ is
A.
zero
B.
$$\left( {{m_1} + {m_2}} \right)g{t_0}$$
C.
$$\frac{1}{2}\left( {{m_1} + {m_2}} \right)g{t_0}$$
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