Discussion Forum



1)

A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at O  and constant angular speed $\omega$.If the angular momentum  of the system, calculated  about O and P  are denoted  by $L_{o}$ and $L_{p}$ respectively, then 

 24112021759_m8.PNG


A) $L_{o}$ and $L_{p}$ do not vary with time

B) $L_{o}$ varies with time while $L_{p}$ remains constant

C) $L_{0}$ remains constant while $L_{p}$ varies with time

D) $L_{o}$ and $L_{p}$ both vary with time

Answer:

Option C

Explanation:

Angular momentum of a particle about a
point is given by :

$L = r \times p =m (r \times v)$

 For $L_{0}$

 24112021763_k1.PNG

 

|L|=$(mvr \sin \theta)=m(R \omega)(R) \sin 90^{0}$

 =$mR^{2} \omega$= constant 

Direction of $L_{0}$ is always upwards , Therefore  , complete $L_{0}$  is constant , both  in magnitude as well as direction

For $L_{p}$

 24112021745_k2.PNG

$|L_{p}|=(mvr \sin \theta)$

                     =$(m)(R\omega)(l) \sin90^{0}$

  =$(mRl \omega)$

magnitude  of $L_{p}$ will remain constant but  direction of $L_{p}$ keeps on changing