Question

A sphere of radius 10 cm and mass 25 kg is attached to the lower end of a steel wire of length 5 m and diameter 4 mm which is suspended from the ceiling of a room . The point of support is 521 cm above the floor. When the sphere is set swinging as a simple pendulum, its lowest point just grazes the floor. Calculate the velocity of the ball at its lowest position (Y_(steel) = 2xx10^(11) N//m^(2)).

Answer 1

Answer:

$\huge\underline\mathfrak{Question !!}$

A sphere of radius 10 cm and mass 25 kg is attached to the lower end of a steel wire of length 5 m and diameter 4 mm which is suspended from the ceiling of a room. The point of support is 521 cm above the floor. When the sphere is set swinging as a simple pendulum, its lowest point just grazes the floor. Calculate the velocity of the ball at its lowest position in m/s. (Ysteel=2×1011 N/m2)

$\huge\underline\mathfrak{Answer !!}$

Δl=521−500−20

=1 cm=0.01 m

T−mg=Rmv2

∴T=m(g+Rv2)

T=m(g+lv2)

l=AYTl=(πd2/4)Ym(g+lv2)l

Δl=(πd2/4)Ymgl+mv2

∴V=4mπd2ΔlY−gl

V=4×25(3.14)(4×10−3)2(0.01)(2×1011)−9.8×5

V=31 m/s