A 5.0 kg ball is whirled on a 1.2 m string so that the ball moves in uniform circular motion in a horizontal plane. The centripetal force is 22 N. What is the tangential speed of the ball?
Answer:
2.3 m/s
Fc=22N
m= 5.0kg
r= 1.2m
vt= √(Fc* r /m)
√(22N *1.2 / 5.0kg) = 2.3 m/s

Respuesta :

Answer: The tangential speed of the ball is 2.3 m/s.

Explanation:

Mass of the ball ,M= 5.0 kg

Length of the string to which the ball is tied = r = 1.2 m

Tangential speed of the ball - v

Centripetal force acting = 22 N

[tex]F_{c}=\frac{M\times (^2)}{r}[/tex]

[tex]22 N=\frac{5.0 kg\times v^2}{1.2 m}[/tex]

[tex]v^2=\frac{22N\times 1.2 m}{5.0 kg}=5.28 m^2/s^2[/tex]

[tex]v=\sqrt{5.28 m^2/s^2}=2.297 m/s\approx 2.3 m/s[/tex]

The tangential speed of the ball is 2.3 m/s.