Question

Gravitational field at the center of the semicircle formed by a thin wire AB of mass M and length is

Answer 1

Explanation:

Assume a small element of mass dM at angular position θ from x axis.

We get dM=

π

M

dθ

Gravitational field due to small element of mass dE=

l

2

GdM

=

πl

2

GM

dθ

Components of dE along x-axis cancel out each other while that along y-axis add up.

Total gravitational field E

net

=∫

0

2

π

2dEsinθdθ (along y axis)

E

net

= ∫

0

2

π

πl

2

2GM

sinθdθ

E

net

=

πl

2

2GM

(−cosθ)

∣

∣

∣

∣

∣

0

2

π

E

net

=

πl

2

2GM

(−0−(−1))

∣

∣

∣

∣

∣

0

2

π

E

net

=

πl

2

2GM

(along y axis)

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