Question 723: Centre of Mass and Rotational Motion

Question

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 xy-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 $${\overrightarrow L _O}$$ and $${\overrightarrow L _P}$$, respectively, then

Accepted Answer

Angular momentum of a particle about  a point is given by:
L = r $$	imes$$ p = m (r $$	imes$$ v)
For L
O
$$\left| L ight| = (mvr\sin 	heta ) = m(R\omega )(R)\sin 90^\circ $$
$$ = m{R^2}\omega $$ = constant
Direction of L
O
is always upwards. Therefore, complete L
O
is constant, both in magnitude as well as direction.
For L
P
$$\left| {{L_P}} ight| = (mvr\sin 	heta ) = m(R\omega )(l)\sin 90^\circ $$
$$ = (mRl\omega )$$
Magnitude of L
P
will remain constant but direction of L
P
keeps on changing.

Answer

$${\overrightarrow L _O}$$ and $${\overrightarrow L _P}$$ do not vary with time.

Answer

$${\overrightarrow L _O}$$ varies with time while $${\overrightarrow L _P}$$ remains constant.

Answer

$${\overrightarrow L _O}$$ remains constant while $${\overrightarrow L _P}$$ varies with time.

Answer

$${\overrightarrow L _O}$$ and $${\overrightarrow L _P}$$ both vary with time.

Centre of Mass and Rotational Motion

Choose the correct answer: