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Question: A particle is moving on a circular path of radius \[10\;{\rm{m}}\] with uniform speed of \[4\;{\rm{m...

A particle is moving on a circular path of radius 10  m10\;{\rm{m}} with uniform speed of 4  ms−14\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}. The magnitude of change in velocity of particle when it completes a semi- circular path is
1. 4  ms−14\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}
2. 8  ms−18\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}
3. 10  ms−110\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}
4. Zero

Explanation

Solution

The above problem can be resolved using the concepts and application of the change of magnitude of velocity along the semi-circular path. In this problem, the basic concept being applied here is that the change in the magnitude of velocity is equal to the difference in the magnitude of velocity at initial and the final point.

Complete step by step answer:
The radius of the circular path is, r=10  mr = 10\;{\rm{m}}.
The speed of travelling through circular path is, v=4  ms−1v = 4\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}
When a particle is supposed to move in a circular path, such that we need to obtain the magnitude of change in velocity of the particle for the complete semi- circular path.
Then we know that while taking the semi-circular path under the observation, the particle will move towards the upward direction first and then it will move to downwards.
Let during the upward travel of a semi-circular path, the magnitude of velocity be v and for the upward direction its sign convention is positive.
Then, expression is,
v1=v{v_1} = v………………………….. (1)
And the magnitude of velocity at downward is taken negative by sign convention.
Then expression is,
v2=−v{v_2} = - v………………………………..(2)
The magnitude of change in velocity along the semi-circular path, using equation 1 and 2 is given as,
Δv=v1−v2\Delta v = {v_1} - {v_2}
Solve by substituting the values in the above equation as,

\Delta v = {v_1} - {v_2}\\\ \Delta v = v - \left( { - v} \right)\\\ \Delta v = 2v \end{array}$$ Further solving as, $$\begin{array}{l} \Delta v = 2v\\\ \Rightarrow \Delta v = 2 \times \left( {{\rm{4}}\;{\rm{m/s}}} \right)\\\ \Rightarrow \Delta v = 8\;{\rm{m/s}} \end{array}$$ Therefore, the magnitude of change in velocity along the semi- circular path is $$8\;{\rm{m/s}}$$ **So, the correct answer is “Option B”.** **Note:** To resolve the given problem, it is sufficient to understand the meaning of the angular velocity as well as the change in the speed, while going from one point to the final point.