Chapter 6, Problem 86A

(a)

To determine

The tension in the string.

Answer to Problem 86A

The tension in the string is $15\text{N}$ .

Explanation of Solution

Given:

The mass of the ball is $m=1.45\text{kg}$ .

The length of string is $l=0.80\text{m}$ .

The angle made by the string with the vertical is $\theta =41°$ .

Formula used:

The expression for the centripetal force acting on an object moving in a circular path is given by,

  $F_{\text{c}}=\frac{m{v}^2}{r}$

Here, $m$ is the mass of the object, $v$ is the velocity of the object, and $r$ is the radius of the circular path.

The expression for the gravitational force acting on the body is,

  $F_{\text{g}}=ma$

Here, $m$ is the mass of the object and $a$ is the acceleration of the object.

Calculation:

The swing of the ball in the circular path is shown below.

  

The gravitational force acting on the ball is denoted by $F_{\text{g}}$ .

The centripetal force acting on the ball is denoted by $F_{\text{c}}$ .

Consider the force in the string is denoted by $F_{\text{T}}$ .

The vertical component $\left(F_{\text{VT}}\right)$ of the tension force in the string is,

  $F_{\text{VT}}=F_{\text{T}}\cos \theta$

The vertical component $\left(F_{\text{VT}}\right)$ of the tension in the string balances the gravitational force $\left(F_{\text{g}}\right)$ acting on the ball.

  $\begin{array}{l}{F}_{\text{VT}}=F_{\text{g}}\\ F_{\text{T}}\cos \theta =mg\\ F_{\text{T}}=\frac{mg}{\cos \theta }\end{array}$

Substitute the values for $m=1.45\text{kg}$ , $g=9.8{\text{m/s}}^{\text{2}}$ and $\theta =14.0°$

  $\begin{array}{c}{F}_{\text{T}}=\frac{(1.45\text{kg})(9.8{\text{m/s}}^{\text{2}})}{\cos \left({14.0}^{\circ }\right)}\\ =\frac{14.21\text{kg}{\text{.m/s}}^{\text{2}}}{0.97}\\ =15\text{N}\end{array}$

Conclusion:

Therefore, the tension in the string is $15\text{N}$ .

(b)

To determine

The speed of the ball.

Answer to Problem 86A

The speed of the ball is $0.69\text{m/s}$ .

Explanation of Solution

Given:

The mass of the ball is $m=1.45\text{kg}$ .

The length of string is $l=0.80\text{m}$ .

The angle made by the string with the vertical is $\theta =41°$ .

Formula used:

The expression for the centripetal force acting on an object moving in a circular path is given by,

  $F_{\text{c}}=\frac{m{v}^2}{r}$

Here, $m$ is the mass of the object, $v$ is the velocity of the object, and $r$ is the radius of the circular path.

The expression for the gravitational force acting on the body as follows:

  $F_{\text{c}}=ma$

Here, $m$ is the mass of the object and $a$ is the acceleration of the object.

Calculation:

The swing of the ball in the circular path is shown below.

  

The gravitational force acting on the ball is denoted by $F_{\text{g}}$ .The centripetal force acting on the ball is denoted by $F_{\text{c}}$ .Consider the force in the string is denoted by $F_{\text{T}}$ .

The radius $\left(r\right)$ of the string is,

  $\begin{array}{l}r=l\sin \theta \\ r=0.80\times \sin \left(14°\right)\\ r=0.1935\text{m}\end{array}$

The horizontal component $\left(F_{\text{HT}}\right)$ of the tension force in the string is

  $F_{\text{HT}}=F_{\text{T}}\sin \theta$

The horizontal component $\left(F_{\text{HT}}\right)$ of the tension in the string balances the centripetal force $\left(F_{\text{c}}\right)$ acting on the ball.

  $\begin{array}{l}{F}_{\text{HT}}=F_{\text{c}}\\ F_{\text{T}}\sin \theta =\frac{m{v}^2}{r}\\ F_{\text{T}}\sin \theta =\frac{m{v}^2}{r}\end{array}$

From part (a), substitute $\left(\frac{mg}{\cos \theta }\right)$ for $F_{\text{T}}$ in the above equation,

  $\begin{array}{l}\left(\frac{mg}{\cos \theta }\right)\sin \theta =\frac{m{v}^2}{r}\\ mg\tan \theta =\frac{m{v}^2}{r}\\ v^2=gr\tan \theta \\ v=\sqrt{gr\tan \theta }\end{array}$

Substitute the values for $r=0.1935\text{m}$ , $g=9.80{\text{m/s}}^{\text{2}}$ and $\theta =14.0°$ .

  $\begin{array}{c}v=\sqrt{(0.1935)(9.8)\tan \left(14.0°\right)}\\ =\sqrt{0.4728}\\ =0.69\text{m/s}\end{array}$

Conclusion:

Therefore, the speed of the ball is $0.69\text{m/s}$ .