Download L m T L/2 L = 0.8m m = 2kg R T A T L θ L/2 L/2 L/2cosθ T v mgsinθ h

L
m
T
L/2
R
L = 0.8m
m = 2kg
T
A
T
θ
L/2
L
L/2
L/2cosθ
T
v
mgsinθ
h
mgcosθ
mg
At the top position the pendulum bob has only potential energy equal to mgL (with respect to its
lowest position)
At the lowest position whole of the potential energy has been converted into kinetic energy. Hence,
KE at the lowest position = mgL ……(1)
As the bob rises on the other side, moving in a circular path of radius L/2, its KE is gradually
converted into PE. PE for any angle θ is given by : mgh = mg(L/2 – L/2 cosθ) = mgL(1-cosθ)/2
KE at angle θ = Total KE at the lowest position – PE at angle θ = mgL - mgL(1-cosθ)/2
KE at angle θ = mgL[1 - (1-cosθ)/2] = mgL(1+cosθ)/2
Or ½ mv2 = mgL(1+cosθ)/2
……(2) where v is the tangential velocity of the bob at angle θ
For moving in a circular path of radius R, the bob needs centripetal force equal to mv2/R. Here, R =
L/2. This centripetal force acts towards the centre of the circular path i.e. pin A. the weight of the
bob can be resolved into components mgsinθ and mgcosθ. Tension T in the string acts as shown in
the fig.. Net force acting on the bob towards pin A = (T - mgcosθ). This force provides the required
centripetal force. Hence,
(T - mgcosθ) = mv2/(L/2) = 2mv2/L
Substituting for mv2 from (2) we get : (T - mgcosθ) = 2mg(1+cosθ) = 2mg + 2mgcosθ
Or T = 2mg + 3mgcosθ = mg(2 + 3cosθ)
Substituting values we get : T = 2 x 9.8 (2 + 3cos30O) = 90 Newtons
To determine the resultant force R acting on the pin A, we apply parallelogram law of addition of
forces as follows :
___________________________
___________________
2
2
O
O
Magnitude R = √T + T + 2TxTxcos(180 – 30 ) = √2x902 + 2x902 cos150O = 46.6 N