Question 347: Work, Energy and Power & Circular Motion

Question

The work done on a particle of mass m by a force
$$$K\left[ {{x \over {{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}}\widehat i + {y \over {{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}}\widehat j} \right]$$$
(K being a constant of appropriate dimensions), when the particle is taken from the point $$\left( {a,0} \right)$$ to the point $$\left( {0,a} \right)$$ along a circular path of radius a about the origin in the x-y plane is

Accepted Answer

The work done on a particle of mass
m
by a force
$$ \vec{F}=K\left[\frac{x}{\left(x^2+y^2ight)^{3/2}} \hat{i}+\frac{y}{\left(x^2+y^2ight)^{3/2}} \hat{j}ight] $$
(with
K
being a constant with appropriate dimensions), when the particle is taken from the point
(a,0)
to the point
(0,a)
along a circular path of radius
a
about the origin in the x-y plane is:
Let $ \vec{r}=x \hat{i}+y \hat{j} $
Therefore, $ d \vec{r}=d x \hat{i}+d y \hat{j} $ and $ r=\sqrt{x^2+y^2} $
Work done by the force in moving a particle from A to B is
$$\begin{aligned}  W & =\int_{r_A}^{r_B} \vec{F} \cdot d \vec{r} \\  & =\int_{r_A}^{r_B} K \left[\frac{x}{(x^2+y^2)^{3/2}} \hat{i}+\frac{y}{(x^2+y^2)^{3/2}} \hat{j}ight] \cdot (d x \hat{i}+d y \hat{j}) \\  & =K \int_{r_A}^{r_B} \frac{x d x}{(x^2+y^2)^{3/2}} + \frac{y d y}{(x^2+y^2)^{3/2}} \\  & =K \int_{r_A}^{r_B} \frac{1}{(x^2+y^2)^{3/2}} \left[d\left(\frac{x^2}{2}ight) + d\left(\frac{y^2}{2}ight)ight] \\  & =K \int_{r_A}^{r_B} \frac{1}{2(x^2 + y^2)^{3/2}} d(x^2 + y^2) \\  & =K \int_{r_A}^{r_B} \frac{1}{2 r^3} d(r^2) = K \int_{r_A}^{r_B} \frac{2 r d r}{2 r^3} = K \int_{r_A}^{r_B} \frac{d r}{r^2} \quad (\because r=\sqrt{x^2 + y^2}) \\  & =K\left[-\frac{1}{r}ight]_{r_A}^{r_B} = K\left[\frac{1}{r_A} - \frac{1}{r_B}ight]  \end{aligned}$$
Here, $ r_A = a $ and $ r_B = a $
Therefore, $ W = 0 $

Answer

$${{2K\pi } \over a}$$

Answer

$${{K\pi } \over a}$$

Answer

$${{K\pi } \over {2a}}$$

Answer

0

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