Respuesta :
Answer:
minimum speed at the bottom, [tex]v_{b(min)}[/tex] = 4.84 m/s
Given:
Length of the rod, l = 59.7 cm =0.597 m
Solution:
According to the conditions given in the question, the gravitation force is the only conservative force acting.
Now by the law of conservation of energy for the rod:
total energy at the bottom end = total energy at the top end
Since, total energy is the sum of Kinetic Energy (KE) and Potential energy (PE)
and we know that PE is maximum at the top and minimum or zero at the bottom whereas KE is min at the top or zero and max at the bottom as:
KE = [tex]\frac{1}{2}m\times v^{2}[/tex]
PE = mgl
As the rod goes over ther top of the circle, the length at the top becomes twice and hence
[tex]PE_{top} mg\times 2l[/tex] = 2mgl
Therefore,
[tex]KE_{bottom} + PE_{bottom} = KE_{top} + PE_{top}[/tex]
[tex]\frac{1}{2}m\times v_{b(min)}^{2} + 0 = 0 + 2mgl[/tex]
⇒ [tex]v^{2} = 4gl[/tex]
⇒ [tex]v = \sqrt 4gl[/tex]
⇒ [tex]v = \sqrt 4\times 9.8\times 0.597 = 4.84 m/s[/tex]
Therefore, minimum speed at the bottom is [tex]v_{b(min)}[/tex] = 4.84 m/s
