cristianosubo wrote:
An object with mass m is spinning in uniform circular motion in a circle perpendicular to the ground with radius r at a circular velocity of 2 m/s. At the height of its spin, the forces acting on it, f, can be calculated by summing the force due to change in direction and the force due to gravity. The force due to gravity is the mass of the object multiplied by 9.8 m/\(s^2\). The force due to change in direction is the product of the square of the velocity of the object, the mass of the object, and the reciprocal of the radius. What is m in terms of r and f?
A) \(\frac{f.r}{4 + 9.8r}\)
B) \(\frac{f}{4r + 9.8}\)
C) \(\frac{f.r}{4r + 9.8}\)
D) \(\frac{4r + 9.8}{f}\)
E) \(\frac{(f.r) - 4}{9.8r}\)
Given;
Mass of object \(= m\)
Velocity of the object \(= 2\)
Radius \(= r\)
Force \("f"\) \(=\) Force due to change in direction \(+\) Force due to gravity
Force due to change in direction \(=\) \((\)Square of velocity of the object \(*\) mass of the object \()/\)Radius \(=> \frac{2^2*m}{r} = \frac{4m}{r}\)
Force due to gravity \(=\) mass of object \(* 9.8 => m*9.8\)
Therefore; \(f = \frac{4m}{r} + m*9.8 = m(\frac{4}{r}+9.8)\)
\(f = m (\frac{4 + 9.8*r}{r})\)
\(m = \frac{f*r}{4+9.8r}\)
Answer A