Chapter 5, Problem 1SP

(a)

To determine

The centripetal acceleration of the ball.

Answer to Problem 1SP

The centripetal acceleration of the ball is $26.67{\text{ m/s}}^2$.

Explanation of Solution

Given Info: The radius of the circle is $0.60\text{ m}$ and the speed of the ball is $4.0\text{ m/s}$.

Write the equation for the centripetal acceleration.

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

Here,

$a_c$ is the centripetal acceleration of the ball

$v$ is the speed of the ball

$r$ is the radius of the circular path

Substitute $4.0\text{ m/s}$ for $v$ and $0.60\text{ m}$ for $r$ in the above equation to find $a_c$.

$\begin{array}{c}{a}_c=\frac{{\left(4.0\text{ m/s}\right)}^2}{0.60\text{ m}}\\ =26.67{\text{ m/s}}^2\end{array}$

Conclusion:

Thus the centripetal acceleration of the ball is $26.67{\text{ m/s}}^2$.

(b)

To determine

The magnitude of the horizontal component of the tension in the string required to produce the centripetal acceleration.

Answer to Problem 1SP

The magnitude of the horizontal component of the tension in the string required to produce the centripetal acceleration is $5.34\text{ N}$.

Explanation of Solution

Given Info: The mass of the ball is $0.20\text{ kg}$.

The horizontal component of tension in the string provides the centripetal force and the magnitude of horizontal component of tension is equal to the magnitude of the centripetal force.

Write the equation for centripetal force.

$F_c=m{a}_c$

Here,

$F_c$ is the magnitude of the centripetal force

$m$ is the mass of the ball

$a_c$ is the centripetal acceleration of the ball

Substitute $0.20\text{ kg}$ for $m$ and $26.67{\text{ m/s}}^2$ for $a_c$ in the above equation to find $F_c$.

$\begin{array}{c}{F}_c=\left(0.20\text{ kg}\right)\left(26.67{\text{ m/s}}^2\right)\\ =5.34\text{ N}\end{array}$

Conclusion:

Thus the magnitude of the horizontal component of the tension in the string required to produce the centripetal acceleration is $5.34\text{ N}$.

(c)

To determine

The magnitude of the vertical component of the tension required to support the weight of the ball.

Answer to Problem 1SP

The magnitude of the vertical component of the tension required to support the weight of the ball is $1.96\text{ N}$.

Explanation of Solution

Given Info: The mass of the ball is $0.20\text{ kg}$.

In the vertical direction there is no any motion. This means that the vertical component of tension supports the weight of the ball so that vertical component of tension in the string is equal to the weight of the ball. Thus, total force in the vertical direction is zero.

Write the equation for the weight of the ball.

$W=mg$

Here,

$W$ is the weight of the ball

$g$ is the acceleration due to gravity

The value of $g$ is $9.8{\text{ m/s}}^2$.

Substitute $0.20\text{ kg}$ for $m$ and $9.8{\text{ m/s}}^2$ for $g$ in the above equation to find $W$.

$\begin{array}{c}W=\left(0.20\text{ kg}\right)\left(9.8{\text{ m/s}}^2\right)\\ =1.96\text{ N}\end{array}$

Conclusion:

Thus the magnitude of the vertical component of the tension required to support the weight of the ball is $1.96\text{ N}$.

(d)

To determine

The vector diagram showing the two components of tension in the string and to estimate the magnitude of the total tension from the diagram.

Answer to Problem 1SP

The vector diagram showing the two components of tension in the string is given in figure 1.

Explanation of Solution

The following figure gives the component of tension in the string.

Figure 1

Here, $\text{T}$ is the tension in the string, ${\text{T}}_h$ is the horizontal component of tension and ${\text{T}}_V$ is the vertical component of tension.

The magnitude of tension $\text{T}$ in the above diagram can be calculated using triangle rule of vector addition. The following figure gives the modified vector diagram to get $\text{T}$.

Figure 2

If base of a triangle gives the horizontal component of tension and height of the triangle gives the vertical component of tension, then the length of the hypotenuse of the triangle will give the magnitude of the total tension in the string.

Write the equation for the tension in the string.

$\text{T}=\sqrt{{\text{T}}_h^2+{\text{T}}_V^2}$

Here,

$\text{T}$ is the tension in string

${\text{T}}_h$ is the horizontal component of tension in the string

${\text{T}}_V$ is the vertical component of the tension in the string

The horizontal component of the tension is equal to centripetal force and vertical component of tension in the string is equal to weight of the body.

Therefore, Substitute $5.34\text{ N}$ for ${\text{T}}_h$ and $1.96\text{N}$ for ${\text{T}}_V$ in the above equation to find the magnitude of the total tension in the string.

$\begin{array}{c}\text{T}=\sqrt{{\left(5.34\right)}^2+{\left(1.96\text{ N}\right)}^2}\\ =5.6\text{ N}\end{array}$

Conclusion:

The vector diagram showing the two components of tension in the string is plotted in figure 1 and the total tension in the string is $5.6\text{ N}$.