Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and length l is
Correct Answer :
2πGm/l² along y axis
Solution :
The correct option is 2πGm/l² along y axis.
Let us derive the gravitational field at the center of the semicircle step-by-step.
Consider a thin wire of mass m and length l bent into a semicircle. Let the center of the semicircle be at the origin (0,0) of a Cartesian coordinate system, and let the semicircle lie in the upper half-plane (y > 0) symmetric about the y-axis. The ends of the wire lie on the x-axis.
The length of the semicircular wire of radius R is:
Therefore, the radius R of the semicircle is:
The mass per unit length (linear mass density λ) of the wire is:
Let us consider a small element of the wire subtending an angle at the center, located at an angle with respect to the positive x-axis.
The length of this small element is , and its mass is:
The gravitational field at the center due to this mass element is directed towards the element (along the radial direction at angle ):
Due to the symmetry of the semicircle about the y-axis, the horizontal components of the gravitational field () from opposite halves of the semicircle cancel each other out. Thus, the net gravitational field along the x-axis is zero:
The vertical components () reinforce each other and point in the positive y-direction (towards the wire since gravity is attractive).
Integrating from to :
Substitute the values of and into the equation:
Therefore, the net gravitational field at the center of the semicircle is 2πGm/l² along y axis.
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