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Sagot :
For a mass m1 on the frictionless table, the speed m1 must rotate a with radius r if m2 is to remain hanging at rest is mathematically given as
[tex]v=\sqrt{m1/m2*xg}[/tex]
What speed must m1 rotate in a circle of radius r if m2 is to remain hanging at rest?
Generally, the equation for the Force is mathematically given as
Fc=Fw
Therefore
[tex]\frac{m1v1}{x}=m2g[/tex]
[tex]v=\sqrt{m1/m2*xg}[/tex]
In conclusion, if m2 is to remain hanging at rest the speed of ratio of m1 is calcuylated using
[tex]v=\sqrt{m1/m2*xg}[/tex]
Read more about Speed
https://brainly.com/question/4931057
The value of the velocity for the mass m₁ on the frictionless table will be [tex]\rm v = \sqrt{\frac{m_1}{m_2 xg} }[/tex]. Velocity is a time-based component. Its unit is m/sec.
What is velocity?
The change of distance with respect to time is defined as speed. Speed is a scalar quantity.
The given data in the problem is
m is the mass
g is the acceleration of free fall =10m/sec²
v is the velocity
From the balancing equation of force the centripetal force is equal to the weight;
[tex]\rm F_c= m_2 g \\\\ \rm \frac{m_1v_1^2}{x} = m_2 g \\\\ v= \sqrt{\frac{m_1}{m_2xg} } \\\\[/tex]
Hence the value of the velocity for the mass m₁ on the frictionless table will be [tex]\rm v = \sqrt{\frac{m_1}{m_2 xg} }[/tex].
To learn more about the velocity refer to the link;
https://brainly.com/question/862972
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