Question

Find the magnitude of the gravitational force this wire exerts on a point with mass m placed at the center of curvature of the semicircle.

Answer 1

Answer:

$F = 2\pi \dfrac{mMG}{L^2}}$

Explanation:

Assuming given,

Mass of wire be M

length of wire be L

small mass at center is = m

Radius of the wire be equal to = R = L/π

mass of small element of the wire

$dM = \dfrac{M}{L}Rd\theta$

All the force are acting along y- direction

so, for force calculation

$F = \int \dfrac{mdMG}{R^2} sin\theta$

$F = \int_0^{\pi} \dfrac{m\dfrac{M}{L}RG}{R^2} sin\theta d\theta$

$F = \int_0^{\pi} \dfrac{mMG}{L\dfrac{L}{\pi}} sin\theta d\theta$

$F = \pi \dfrac{mMG}{L^2}}\int_0^{\pi}sin\theta d\theta$

$F = \pi \dfrac{mMG}{L^2}}(-cos \theta)_0^{\pi}$

$F = 2\pi \dfrac{mMG}{L^2}}$

Answer 2

The magnitude of the gravitational force is : $2\pi \frac{mMG}{L^{2} }$

Given that :

mass of wire = M

length of wire = L

small mass at center = m

radius of wire = L / $\pi$

Determine the magnitude of the gravitational force exerted by the wire on a point

First step : express the mass of small element of wire

dM = $\frac{M}{L} Rd$∅

Since all forces act in the vertical direction the magnitude of the force exerted will be

F = $\int\limits^\pi _o {\frac{m\frac{M}{L} RG}{R^2} } \, sin\beta d\beta$

$F = \pi \frac{mMG}{L^2} \int\limits^\pi _0 {x} \, sin\beta d\beta$

Resolving equation above

Therefore F = $2\pi \frac{mMG}{L^{2} }$

Hence we can conclude that The magnitude of the gravitational force is : $2\pi \frac{mMG}{L^{2} }$

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