Question 379: Centre of Mass and Rotational Motion

Question

Let $$\overrightarrow F $$  be the force acting on a particle having position vector $$\overrightarrow r ,$$ and $$\overrightarrow 	au  $$ be the torque of this force about the origin. Then

Accepted Answer

As we know $$\overrightarrow 	au   = \overrightarrow r  	imes \overrightarrow F $$
So the angle between $$\overrightarrow 	au  $$ and $$\overrightarrow r $$ is $${90^ \circ }$$ and the  angle between $$\overrightarrow t $$ and $$\overrightarrow F $$ is also $${90^ \circ }.$$
We also know that the dot product of two vectors which have an angle of $${90^ \circ }$$  between them is zero.
$$	herefore$$  $$\overrightarrow {r.} \vec 	au = 0{\mkern 1mu} $$ and $$\overrightarrow {F.} \overrightarrow 	au  = 0\,\,$$
Therefore $$(d)$$ is the correct option.

Answer

$$\overrightarrow {r.} \overrightarrow 	au  = 0\,\,$$ and $$\overrightarrow {F.} \overrightarrow 	au  
e 0\,\,$$

Answer

$$\overrightarrow {r.} \vec 	au 
e 0{\mkern 1mu} {\mkern 1mu} $$ and $$\overrightarrow {F.} \overrightarrow 	au  = 0\,\,$$

Answer

$$\overrightarrow {r.} \vec 	au 
e 0{\mkern 1mu} $$ and $$\overrightarrow {F.} \overrightarrow 	au  
e 0$$

Answer

$$\overrightarrow {r.} \vec 	au = 0{\mkern 1mu} $$ and $$\overrightarrow {F.} \overrightarrow 	au  = 0\,\,$$

Centre of Mass and Rotational Motion

Let $\overrightarrow F be the force acting on a particle having position vector \overrightarrow r , and \overrightarrow \tau $ be the torque of this force about the origin. Then

Choose the correct answer: