Question

A heavy particle hanging from a string of length l is projected horizontally with speed sqrtgl. Find the speed of the particle at the point where the tension in the string equals weight of the particle.

Answer 1

Answer:

Let T=mg

h=l(1−cosθ)

using energy conservation

2

1

m(U

2

−V

2

)=mgh

U=gl, V= speed of particle at B

V

2

=V

2

−2gh

T−mgcosθ=

l

mV

2

mg−mgcosθ=

l

mV

2

V

2

=gl(1−cosθ)

gl(1−cosθ)=gl−2gl(1−cosθ)

cosθ=

3

2

∴V=

3

gl

Explanation:

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