A pendulum consists of a bob of mass m=0.1kg and a massless inextensible string of length L=1.0m. It is suspended from a fixed point at height H=0.9m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse P=0.2kg−m/s is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is Jkg−m2/s. The kinetic energy of the pendulum just after the lift-off is K Joules.
The value of K is ____.
Answer: 0.16
Solution
From the geometry, when the string becomes taut:
cosθ=LH=1.00.9=0.9
Just after the string becomes taut, the component of velocity along the string becomes zero (impulsive tension). Only the perpendicular component remains:
v⊥=vcosθ=2×0.9=1.8 m/s
Alternatively, using angular momentum conservation:
J=mv⊥L
0.18=0.1×v⊥×1.0
v⊥=1.8 m/s
Kinetic energy just after lift-off:
K=21mv⊥2=21×0.1×(1.8)2=0.05×3.24=0.162 J
Rounded to two decimal places: K=0.16 J
JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Numerical answer
