A horizontal force of 310 N is exerted on a 2.0-kg ball as it rotates (at arm’s length) uniformly in a horizontal circle of radius 0.90 m. Calculate the speed of the ball.

As the ball is rotated along the horizontal circle, it experiences a centripetal force acting towards the centre of the horizontal circle.

The magnitude of the centripetal force depends on the speed of the ball and the radius of the horizontal circle.

The given data can be listed below as,

• The magnitude of horizontal force on ball is, $F_{\mathrm{c}}=310\mathrm{N}$.
• The mass of ball is, $m=2\mathrm{kg}$.
• The radius of horizontal circle is, $r=0.90\mathrm{m}$.

The centripetal force on the ball can be expressed as,

$\begin{array}{c}{F}_{\mathrm{c}}=m{a}_{\mathrm{c}}\\ F_{\mathrm{c}}=\frac{m{v}^2}{r}\\ v=\sqrt{\frac{F_{\mathrm{c}}\times r}{m}}\end{array}$

Here, v is the speed of the ball, r is the radius of the horizontal circle,$a_{\mathrm{c}}$is the centripetal acceleration of the ball.

Substitute the values in the above equation.

$\begin{array}{c}v=\sqrt{\frac{310\mathrm{N}\times 0.90\mathrm{m}\left(\frac{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)}{2\mathrm{kg}}}\\ v=\sqrt{139.5}\mathrm{m}/\mathrm{s}\\ v=11.81\mathrm{m}/\mathrm{s}\end{array}$

Thus, the speed of the ball is $11.81\mathrm{m}/\mathrm{s}$.