Athin uniform rod (mass = 0.53 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. the rod swings with a period of 1.4 s and an angular amplitude of 3.9°.
(a) what is the length of the rod?
(b) what is the maximum kinetic energy of the rod as it swings?
(a) L = 0·73 m
(b) 4·39 × J
(a) From the figure, consider the torque about the point where the rod is attached because if we consider another point then there will be hinge forces acting on the rod at the point of attachment
Let L m be the length of the rod and β be the angle between the rod and the vertical
Let α be the angular acceleration of the rod
As the force of gravity acts at the centre so from the figure, the torque about the point of attachment will be 0·53 × g ×(L ÷ 2) ×sinβ
Assuming that the value of amplitude of this oscillation to be small
As torque = moment of inertia × angular acceleration
0·53 × g ×(L ÷ 2) ×sinβ = ((0·53 × L²) ÷ 3) × α (∵ moment of inertia of the rod from the point of attachment)For small oscillations, α = ω² × β
After substituting the value of α and solving we get
ω = √((3 × g) ÷ (2 × L))
Time period = (2 × π) ÷ ω = (2 × π) ÷ √((3 × g) ÷ (2 × L))
∴ (2 × π) ÷ √((3 × g) ÷ (2 × L)) = 1·4
Substituting the value of g as 9·8 m/s² and solving we get
L = 0·73 m
(b) At the maximum amplitude condition the velocity will be 0 and potential energy will be maximum and maximum kinetic energy will be attained at the lowest point and hinge forces will not do work as the point of attachment is not moving
∴ Taking the reference for finding the potential energy as the lowest pointMaximum potential energy = Maximum kinetic energy As total energy is constant, since there is no dissipative force
Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 (∵ increment in height is (L × (1 - cosβ)) ÷ 2
∴ Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 After substituting the value we get
Maximum potential energy = 4·39 × J
∴ Maximum kinetic energy = 4·39 × J
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