
The relation between the linear velocity and the angular velocity is
(A) $\vec \omega = \vec r \times \vec v$
(B) $\vec v = \vec r \times \vec \omega $
(C) $\vec v = \vec \omega \times \vec f$
(D) $\vec \omega = \vec v \times \vec f$
Answer
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Hint:Motion of the object can take place either linearly or in a circular motion. In case of the linear motion, the displacement, velocity and the acceleration all are specified in linear whereas in the circular motion, all the parameters are specified by the angular term.
Useful formula:
The distance covered by the circular motion is given by
$\partial \vec s = \partial \vec \theta \times \vec r$
Where $s$ is the distance covered by the circular motion, $\partial \vec \theta $ is the distance covered by the object per radius of the circular motion and $\vec r$ is the radius of the circular motion.
Complete step by step solution:
The linear velocity is the rate of the change of the linear distance with that of the time where the angular velocity is considered for the circular displacement. The acceleration that is obtained in linear displacement is linear acceleration and the acceleration that is obtained for the circular motion is the angular acceleration.
Using the formula of the circular motion,
$\partial \vec s = \partial \vec \theta \times \vec r$
Dividing both side of the equation by $\partial t$ , we get
$\dfrac{{\partial \vec s}}{{\partial t}} = \dfrac{{\partial \vec \theta }}{{\partial t}} \times \vec r$
By changing the above equation into the derivative form,
$\dfrac{{d\vec s}}{{dt}} = \dfrac{{d\vec \theta }}{{dt}} \times \vec r$
We know that the rate of change of distance $\left( s \right)$ with time is linear velocity and the rate of the change of the angular distance $\left( \theta \right)$ in the above equation,
$\vec v = \vec \omega \times \vec r$
Thus the option (B) is correct.
Note:The example of the linear motion is the motion of the car in a straight road, man walking on a path, etc. The example of the angular motions are rotating of the disc in a horizontal direction, movement of the mass suspended by the string on a circular motion etc.
Useful formula:
The distance covered by the circular motion is given by
$\partial \vec s = \partial \vec \theta \times \vec r$
Where $s$ is the distance covered by the circular motion, $\partial \vec \theta $ is the distance covered by the object per radius of the circular motion and $\vec r$ is the radius of the circular motion.
Complete step by step solution:
The linear velocity is the rate of the change of the linear distance with that of the time where the angular velocity is considered for the circular displacement. The acceleration that is obtained in linear displacement is linear acceleration and the acceleration that is obtained for the circular motion is the angular acceleration.
Using the formula of the circular motion,
$\partial \vec s = \partial \vec \theta \times \vec r$
Dividing both side of the equation by $\partial t$ , we get
$\dfrac{{\partial \vec s}}{{\partial t}} = \dfrac{{\partial \vec \theta }}{{\partial t}} \times \vec r$
By changing the above equation into the derivative form,
$\dfrac{{d\vec s}}{{dt}} = \dfrac{{d\vec \theta }}{{dt}} \times \vec r$
We know that the rate of change of distance $\left( s \right)$ with time is linear velocity and the rate of the change of the angular distance $\left( \theta \right)$ in the above equation,
$\vec v = \vec \omega \times \vec r$
Thus the option (B) is correct.
Note:The example of the linear motion is the motion of the car in a straight road, man walking on a path, etc. The example of the angular motions are rotating of the disc in a horizontal direction, movement of the mass suspended by the string on a circular motion etc.
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