Question

A ball on the end of a string is whirled around in a horizontal circle of radius$\mathbf{0}\mathbf{.}\mathbf{300}\mathbf{}\mathbf{m}$. The plane of the circle is$\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{m}$above the ground. The string breaks and the ball lands$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$(horizontally) away from the point on the ground directly beneath the ball's location when the string breaks. Find the radial acceleration of the ball during its circular motion.

Short answer

The centripetal acceleration of the ball during the circular motion is $54.4\text{m}/{\text{s}}^2$.

Step-by-step solution

Given data

The radius of the circle is $0.300\text{m}$, the height of the plane above the ground is $1.20\text{m}$and the ball lands $2.00\text{m}$away from the point directly beneath the ball's location.

Formula to calculate horizontal range and time of flight 

The formula to calculate horizontal range of the ball

$\mathit{R}\mathbf{}\mathbf{=}\mathbf{}\mathit{v}\mathit{t}$ ...... (i)

Here, R is the horizontal range, v is the tangential velocity of the ball and t is time of flight of the ball.

The formula to calculate time of fight

$\mathit{t}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{h}}{\mathbf{g}}}$

Here, h is the vertical height from ground and g is the acceleration due to gravity.

Solve for the centripetal acceleration of the ball during its circular motion

Substitute $\sqrt{\frac{2h}{g}}$ for t in equation (i),

$R=v\sqrt{\frac{2h}{g}}$

Substitute 1.20 m for h , 2.00 m for R and $9.81\text{m}/{\text{s}}^2$ for g to find v.

$\begin{array}{rcl}2.00\text{m}& =& v\sqrt{\frac{2\times 1.20\text{m}}{9.81\text{m}/{\text{s}}^2}}\\ v& =& 4.04\text{m}/\text{s}\end{array}$

Write the formula to calculate the radial acceleration

$a=\frac{v^2}{r}$

Here, a is the centripetal acceleration and r is the circular radius.

Substitute $4.04\text{m}/\text{s}$ for v and 0.30 m for r in the above equation to find a.

$\begin{array}{rcl}a& =& \frac{{(4.04\text{m}/\text{s})}^2}{0.30\text{m}}\\ a& =& 54.4\text{m}/{\text{s}}^2\end{array}$

Therefore, the centripetal acceleration of the ball during its circular motion is $54.4\text{m}/{\text{s}}^2$.