Nov 4, 2010

Pendulum.

Q. Pendulum.
Question 
A 0.500 kg pendulum bob passes through the lowest part of its path at a speed of 3.80 m/s.
(a) What is the tension in the pendulum cable at this point if the pendulum is 80.0 cm long?
(b) When the pendulum reaches its highest point, what angle does the cable make with the vertical?
(c) What is the tension in the pendulum cable when the pendulum reaches its highest point?

Answer
(a) 
Two forces act on the pendulum bob: 
(i) its weight, mg, downwards and
(ii) tension, T, in the cable, upwards.

The resultant of these two forces in the upwards direction,
T - mg = mv^2/r, where v = velocity of the bob and r = cable length.
=> T 
= m(g + v^2/r)
= (0.5)[9.8 + (3.8)^2/0.8]
= 13.925 N

(b)
Let h = vertical distance of the highest point from the lowest point, and
θ = angle made by the cable with the vertical at that instant.
Then, h = r - r cosθ = r(1 - cosθ) = 2rsin^2(θ/2)
=> h = (2r)*sin^2(θ/2) ... ( 1 )

When the bob reaches the highest point, its K.E. at the lowest point gets converted to P.E. at the highest point.
So, (1/2)mv^2 = mgh
=> h = v^2/2g ... ( 2 )

From eqns. ( 1 ) and ( 2 ), 
(2r)*sin^2(θ/2) = v^2/2g
=> sin(θ/2) = v / 2√(rg)
=> sin(θ/2) = 3.8 / 2√(0.8)(9.8)
=> sin(θ/2) = 0.68 
=> θ = 85.7°

(c) 
At the highest point, its velocity is zero. So no centripetal force on it. Hence, T should equal the component of mg in its direction.

=> T = mg cos 85.7° = (0.5)(9.8) (0.075)
=> T = 0.37 N.

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