
A rod of length $ l $ is given two velocities $ {v_1} $ and $ {v_2} $ in opposite directions at its two ends at right angles to the length of the rod. The distance of the instantaneous axis of rotation from the end which is given velocity $ {v_1} $ is:
(A) $ \dfrac{{{v_2}l}}{{{v_1} + {v_2}}} $
(B) $ \left( {\dfrac{{{v_1} + {v_2}}}{{{v_1}}}} \right)l $
(C) $ \left( {\dfrac{{{v_1}l}}{{{v_1} + {v_2}}}} \right) $
(D) $ \left( {\dfrac{{{v_1} + {v_2}}}{{{v_2}}}} \right)l $
Answer
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Hint : We need to fix a point of rotation and measure the distances of the ends of the rod from that point. Then we can relate angular velocity with linear velocity to get the answer.
Formula used: The formulae used in the solution are given here.
Angular Velocity $ \omega = \dfrac{v}{r} $ where, $ v $ is the linear velocity and $ r $ is the radius of the circular path.
Complete step by step answer:
Angular velocity is described as the rate of change of angular displacement which specifies the angular speed or rotational speed of an object and the axis about which the object is rotating. It is a vector quantity. The angular velocity and linear velocity is articulated by the formula $ \omega = \dfrac{v}{r} $ where, $ v $ is the linear velocity and $ r $ is the radius of the circular path.
Given that, a rod of length $ l $ is given two velocities $ {v_1} $ and $ {v_2} $ in opposite directions at its two ends at right angles to the length of the rod.
Let the point about which the rod rotates be P. Note that, it is also the intersection point of the lines drawn perpendicular to the velocity of the two ends. Let the distance of the point P to the end where the velocity is $ {v_1} $ be $ X. $ Thus the distance of the point where the velocity is $ {v_2} $ be $ l - X. $
Hence, by using, $ \omega = \dfrac{v}{r} $ , we get the equation,
$ \omega = \dfrac{{{v_1}}}{X} = \dfrac{{{v_2}}}{{l - X}} $
$ \Rightarrow {v_1}l - {v_1}X = X{v_2} $
The distance of the instantaneous axis of rotation from the end which is given velocity $ {v_1} $ is $ X $ .
Thus the value of $ X $ is given by,
$ \Rightarrow X = \dfrac{{{v_1}l}}{{{v_1} + {v_2}}} $
Hence, the correct answer is Option D.
Note:
The amount of change of angular displacement of the particle at a given period of time is called angular velocity. The track of the angular velocity vector is vertical to the plane of rotation, in a direction which is usually indicated by the right-hand rule.
Angular velocity is also, articulated as $ \omega = \dfrac{{d\theta }}{{dt}} $ where, $ d\theta $ is the change in angular displacement and $ dt $ is the change in time.
Formula used: The formulae used in the solution are given here.
Angular Velocity $ \omega = \dfrac{v}{r} $ where, $ v $ is the linear velocity and $ r $ is the radius of the circular path.
Complete step by step answer:
Angular velocity is described as the rate of change of angular displacement which specifies the angular speed or rotational speed of an object and the axis about which the object is rotating. It is a vector quantity. The angular velocity and linear velocity is articulated by the formula $ \omega = \dfrac{v}{r} $ where, $ v $ is the linear velocity and $ r $ is the radius of the circular path.
Given that, a rod of length $ l $ is given two velocities $ {v_1} $ and $ {v_2} $ in opposite directions at its two ends at right angles to the length of the rod.
Let the point about which the rod rotates be P. Note that, it is also the intersection point of the lines drawn perpendicular to the velocity of the two ends. Let the distance of the point P to the end where the velocity is $ {v_1} $ be $ X. $ Thus the distance of the point where the velocity is $ {v_2} $ be $ l - X. $
Hence, by using, $ \omega = \dfrac{v}{r} $ , we get the equation,
$ \omega = \dfrac{{{v_1}}}{X} = \dfrac{{{v_2}}}{{l - X}} $
$ \Rightarrow {v_1}l - {v_1}X = X{v_2} $
The distance of the instantaneous axis of rotation from the end which is given velocity $ {v_1} $ is $ X $ .
Thus the value of $ X $ is given by,
$ \Rightarrow X = \dfrac{{{v_1}l}}{{{v_1} + {v_2}}} $
Hence, the correct answer is Option D.
Note:
The amount of change of angular displacement of the particle at a given period of time is called angular velocity. The track of the angular velocity vector is vertical to the plane of rotation, in a direction which is usually indicated by the right-hand rule.
Angular velocity is also, articulated as $ \omega = \dfrac{{d\theta }}{{dt}} $ where, $ d\theta $ is the change in angular displacement and $ dt $ is the change in time.
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