A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m. If the ball revolves twice every second, what is the tension in the string?

Respuesta :

Answer:

19.74 N

Explanation:

mass of ball (m) = 0.25 kg

radius (r) = 0.5 m

time (t) = 2 revolutions per seconds = 1/2 = 0.5 second per revolution

find the tension in the string

tension (T) = [tex]\frac{mv^{2} }{r}[/tex]

  • where velocity (v) = [tex]\frac{2πr}{t}[/tex]

tension now becomes (T) = [tex]\frac{m}{r} x (\frac{2πr}{t})^{2}[/tex]

tension (T) = [tex]\frac{4π^{2}rm }{t^{2} }[/tex]

  • now substituting the values of mass (m), time (t) and radius (r) into the equation above we have

tension (T) = [tex]\frac{4π^{2}x0.5x0.25}{0.5^{2} }[/tex]

tension (T) =  [tex]2π^{2}[/tex] =   [tex]2x3.142^{2}[/tex] = 19.74 N