
A body of mass 100 g is rotating in a circular path of radius r with constant speed. The work done in one complete revolution is
A. \[\left[ {100r} \right]J\]
B. \[\left[ {\dfrac{r}{{100}}} \right]J\]
C. \[\left[ {\dfrac{{100}}{r}} \right]J\]
D. Zero
Answer
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Hint:The centripetal force acting on the body in uniform circular motion is towards centre and the linear velocity of the body is tangential to the circular path, i.e. the linear displacement of the body is along the tangent to the circular path.
Formula used:
\[W = \left( F \right)\left( {\Delta x} \right)\cos \theta \]
where W is the work done by the force vector \[\overrightarrow F \] for causing a linear displacement of \[\overrightarrow x \] in the body.
Complete step by step solution:
To move the body in a circular path with uniform circular motion there needs a force which should balance the radially outward centrifugal force so that the body remains in the circular path. The radially inward force is called the centripetal force acting along the radius of the circular path. The centripetal force acting on the body in uniform circular motion is towards the centre and the linear velocity of the body is tangential to the circular path, i.e. the linear displacement of the body is along the tangent to the circular path.
When a line joins the point on the circular path at which the tangent is drawn to the centre of the circle then the line drawn is perpendicular to the tangent. So, the angle between the centripetal force and the linear displacement is 90° because the linear displacement is along the tangent to the circular path and the force is along the line joining the point of the tangent to the centre.
If the centripetal force acting on the body is \[F\] and the linear displacement during an interval of time is \[\Delta x\] then using the work formula, the work done on the body during this interval of time will be,
\[W = \left( F \right)\left( {\Delta x} \right)\cos \theta \]
Here, \[\theta = 90^\circ \]
\[W = \left( F \right)\left( {\Delta x} \right)\cos 90^\circ \]
\[\Rightarrow W = \left( F \right)\left( {\Delta x} \right)\left( 0 \right)\]
\[\therefore W = 0\]
Hence, the work done by the force is zero.
Therefore, the correct option is D.
Note: As the given motion is uniform circular motion, the speed of the body is constant, i.e. the kinetic energy of the body is constant. Using the work-energy theorem the net work done is equal to the change in kinetic energy. So, in this case, the work done is zero, as a change in kinetic energy is zero.
Formula used:
\[W = \left( F \right)\left( {\Delta x} \right)\cos \theta \]
where W is the work done by the force vector \[\overrightarrow F \] for causing a linear displacement of \[\overrightarrow x \] in the body.
Complete step by step solution:
To move the body in a circular path with uniform circular motion there needs a force which should balance the radially outward centrifugal force so that the body remains in the circular path. The radially inward force is called the centripetal force acting along the radius of the circular path. The centripetal force acting on the body in uniform circular motion is towards the centre and the linear velocity of the body is tangential to the circular path, i.e. the linear displacement of the body is along the tangent to the circular path.
When a line joins the point on the circular path at which the tangent is drawn to the centre of the circle then the line drawn is perpendicular to the tangent. So, the angle between the centripetal force and the linear displacement is 90° because the linear displacement is along the tangent to the circular path and the force is along the line joining the point of the tangent to the centre.
If the centripetal force acting on the body is \[F\] and the linear displacement during an interval of time is \[\Delta x\] then using the work formula, the work done on the body during this interval of time will be,
\[W = \left( F \right)\left( {\Delta x} \right)\cos \theta \]
Here, \[\theta = 90^\circ \]
\[W = \left( F \right)\left( {\Delta x} \right)\cos 90^\circ \]
\[\Rightarrow W = \left( F \right)\left( {\Delta x} \right)\left( 0 \right)\]
\[\therefore W = 0\]
Hence, the work done by the force is zero.
Therefore, the correct option is D.
Note: As the given motion is uniform circular motion, the speed of the body is constant, i.e. the kinetic energy of the body is constant. Using the work-energy theorem the net work done is equal to the change in kinetic energy. So, in this case, the work done is zero, as a change in kinetic energy is zero.
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