Question

A body of mass in is revolving along a circular path of radius 'R' with uniform speed 'V'. The work done by it in one complete turn is​

Answer 1

mv ²/ r.......

Its your answer

Answer 2

Given:

A body of mass in is revolving along a circular path of radius 'R' with uniform speed v.

To find:

Work done by the body in one complete turn ?

Diagram:

$\setlength{\unitlength}{1.2mm}\begin{picture}(30,30)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)(5.000,38.284)(5.000,30.000)\qbezier(5.000,30.000)(5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(45,30){\vector(-4,0){14}}\put(5,30){\vector(4,0){14}}\put(25,30){\circle*{1}}\put(25,9.9){\vector(0,5){13}}\put(25,50.1){\vector(0,-5){13}}\put(33,32){\sf{ \overrightarrow{F} }}\put(45,31.5){\vector(0,5){20}}\put(5,31){\vector(0,-5){20}}\put(25,10){\vector(4,0){20}}\put(25,50){\vector(-4,0){19}}\put(17,32){\sf{ \overrightarrow{F} }}\put(27,39){\sf{ \overrightarrow{F} }}\put(27,20){\sf{ \overrightarrow{F} }}\put(2,46){\sf{ \overrightarrow{v} }}\put(0,10){\sf{ \overrightarrow{v} }}\put(47,10){\sf{ \overrightarrow{v} }}\put(47,50){\sf{ \overrightarrow{v} }}\end{picture}$

Calculation:

We know that : what can be calculated by the scalar product of force vector and displacement vector.

$\sf \therefore \: W = \vec{F} \: . \: \vec{r}$

$\sf \implies \: W = | \vec{F}| \times | \vec{r}| \times \cos( \theta)$

In a circular motion , $\theta$ = 90°:

$\sf \implies \: W = | \vec{F}| \times | \vec{r}| \times \cos( {90}^{ \circ} )$

$\sf \implies \: W = | \vec{F}| \times | \vec{r}| \times 0$

$\sf \implies \: W = 0 \: joule$

Hence , the net work done by the body while travelling in in a circular trajectory will be zero joule.