Answer :
The tension in the cord when the object is at the lowest point of its swing is 3 mg.
The complete question is in the attachment. The small object moves from point A to point B. According to the law of conservation of energy
KE a + PE a = KE b + PE b
0.5 m₁ v₁² + m₁ g h₁ + 0.5 m₂ v₂² + m g h₂
- At point A
The height = h₁ = r
The velocity = v₁ = 0 - At point B
The height = h₂ = 0
The velocity = v₂
(0.5 × m × 0²) + (m × g × r) = (0.5 × m × v₂²) + (m × g × 0)
0 + mgr = 0.5 mv₂² + 0
v₂² = 2gr
[tex]v_2 \:=\: \sqrt{2gr}[/tex]
The object moves according to uniform circular motion. There is the centripetal force that works in the system. The forces acting on the object are weight force and tension force. According to Newton's second law for circular motion
Fc = mα
T - w = mv²/R
T - mg = mv²/R
- m = the mass
- v = the velocity at the lowest point = v₂
- R = the radius = r
- g = the acceleration due to gravity
T = mv₂²/r + mg
T = (m × 2gr ÷ r) + mg
T = 2mg + mg
T = 3 mg
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