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A simple pendulum has a length of 0.75 m and a bob with a mass of 0.15 kg. The bob is released from an angle of 25 degrees relative to the vertical reference line.

a) Show that the vertical height of the bob when it is released is h= L(1-cos25 )

b) What is the kinetic energy of the bob when the string is at an angle of 9 degrees?

c) What is the speed of the bob at the bottom of the wing? (Neglecting friction and the mass of the string)

a) Lets assume that the vertical height of the bob is H and let the vertical distance from the bob to the point of attachment of the string to the ceiling be X.

X/L = cos 25
X = L cos25
H = L – Lcos25 = L(1-cos25) (SHOWN)

b) The height when the bob is at an angle of 9 degrees, H' = L (1-cos9) = 0.75(1 – cos 9 )

The kinetic energy = mgH – mgH' = 0.15(9.81)[0.75(1 - cos25)] – 0.15(9.81)[0.75(1 - cos9)] = 0.0898 J

c) By conservation of energy,

mgH = 1/2mv^2
gH = 1/2(v^2)
v^2 = 2gH = 1.379
v = 1.17m/s

Hope this is correct ^^

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