Question

A boy whirls a ball on a string 1.0 m long in a horizontal circle at 50 rpm. If the mass of the ball is 0.22 kg

Answer 1

Hi there! :)

We can begin by doing a summation of forces on the ball. In the horizontal direction, we have the force of tension:
$F_{Net} = T$

The tension force results in a centripetal force experienced by the ball. The equation of centripetal force is equivalent to:
$F_C = m\omega^2 r$

$F_C$ = Centripetal force (N)
m = mass of ball (0.22 kg)

ω = angular speed of ball(? rad/sec, must convert rpm to rad/sec)

r = radius/length of string (1.0 m)

We must begin by converting rpm to rad/sec:

$\frac{50 rev}{min} * \frac{2\pi rad}{1 rev}* \frac{1 min}{60 sec} = 5.236 \frac{rad}{s}$

Now, we can set tension equal to the centripetal force and solve. $T = F_{Net} = F_C\\\\T = m\omega^2 r\\\\T = (0.22)(5.236^2)(1) = \boxed{6.031 N}$