First slide

Vertical circular motion

Question

A pendulum with a bob of mass m is suspended from a horizontal platform. The platform is given a horizontal uniform acceleration. The breaking tension in the light string of the pendulum is( 2/√3 ) mg. Find the work done by the extreme tension T on the bob in the first one sec.

Difficult

A

mg² / 6
B

mg³ / 6
C

mg² /3
D

mg² / 4

Solution

Work done, W = (F) (S).
   Force F and displacement are parallel. Here, the bob does not move in y axis.
      ⇒ The work done by the tension T along Y axis is zero.
   Since the bob moves horizontally (along x-axis) with an acceleration, say a, SF_x= ma
⇒ Tsin θ= ma
⇒ a = T sin θ /m .............. (1)
For equilibrium of the bob along Y - axis
SF_y = 0 ⇒ T cos θ = mg .......... (2)
From (1) and (2) ,
$\large a = \left( {\frac{{mg}}{{\cos \theta }}} \right)\left( {\frac{{\sin \theta }}{m}} \right) = g\tan \theta$
⇒ a = g tan θ ................. (3)
Since the breaking strength .
T = 2mg / √3 ⇒ mg / cos θ = 2mg / √3
⇒ θ = 30⁰ , Putting θ = 30⁰ in (3)
We obtain a = g tan 30⁰ = g / √3
Hence required work done = W = F.S.
=(T sin θ) (1/2 at²)
Where T = 2√3 mg ,
θ = 30⁰ , t = 1 s
W = (2mg / √3) sin 30⁰ . (1/2 g/√3 . 1²)
= mg² / 6

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