In Fig. 6-45, a$\mathbf{1}\mathbf{.}\mathbf{34}\mathbf{}\mathbf{kg}$ball is connected by means of two massless strings, each of length$\mathbf{L}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{}\mathbf{m}$, to a vertical, rotating rod. The strings are tied to the rod with separation$\mathbf{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{}\mathbf{m}$and are taut. The tension in the upper string is$\mathbf{35}\mathbf{N}$. What are the

(a) tension in the lower string,

(b) magnitude of the net force${\overrightarrow{\mathbf{F}}}_{\mathbf{net}}$on the ball, and

(c) speed of the ball?

(d) What is the direction of$\mathbf{F}$?

• Mass of ball,$\mathrm{m}=1.34\text{\hspace{0.17em}kg}$.
• Length of each string,$\mathrm{L}=1.70\text{\hspace{0.17em}m}$.
• The separation between two strings tied to rod,$\mathrm{d}=1.70\text{\hspace{0.17em}m}$.
• Tension in the upper string,${\mathrm{T}}_{\mathrm{u}}=35\text{\hspace{0.17em}N}$ .

Using the concept of centripetal force and applying Newton's second law, we can solve the given problem. Note that the tensions in the strings provide the source of centripetal force.

The given system consists of a ball connected by two strings to a rotating rod. The tensions in the strings provide the source of centripetal force.

The free-body diagram for the ball is shown below. ${\overrightarrow{\mathrm{T}}}_{\mathrm{u}}$is the tension exerted by the upper string on the ball, ${\overrightarrow{\mathrm{T}}}_{\mathrm{l}}$is the tension in the lower string, and role="math" localid="1661234901128" $m$is the mass of the ball. Note that the tension in the upper string is greater than the tension in the lower string. It must balance the downward pull of gravity and the force of the lower string.

We take the$+\mathrm{x}$direction to be leftward (toward the center of the circular orbit) and$+\mathrm{y}$upward. Since the magnitude of the acceleration is$\mathrm{a}={\mathrm{v}}^2/\mathrm{R}$, the$\mathrm{x}$component of Newton's second law is,

${\mathrm{T}}_{\mathrm{u}}\text{cos}\mathrm{\theta }+{\mathrm{T}}_{\mathrm{l}}\text{cos}\mathrm{\theta }=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

Where$\mathrm{v}$is the speed of the ball, and$\mathrm{R}$ is the radius of its orbit.
The$\mathrm{y}$ component is,

${\mathrm{T}}_{\mathrm{u}}\text{sin}\mathrm{\theta }-{\mathrm{T}}_{\mathrm{l}}\text{sin}\mathrm{\theta }-\mathrm{mg}=0$

The second equation gives the tension in the lower string:

${\mathrm{T}}_{\mathrm{l}}={\mathrm{T}}_{\mathrm{u}}-\mathrm{mg}/\text{sin}\mathrm{\theta }$.

Since the triangle is equilateral, the angle is$\mathrm{\theta }={30.0}^{\text{o}}$.

Thus,

${\mathrm{T}}_{\mathrm{l}}={\mathrm{T}}_{\mathrm{u}}-\frac{\mathrm{mg}}{\text{sin}\mathrm{\theta }}$

Substitute the values, and we get,

$\begin{array}{l}{\mathrm{T}}_{\mathrm{l}}=35.0\text{\hspace{0.17em}N}-\frac{\left(1.34\text{\hspace{0.17em}kg}\right)(9.80{\text{\hspace{0.17em}m/s}}^{\text{2}})}{\text{sin}{30.0}^{\circ }}\\ {\mathrm{T}}_{\mathrm{l}}=8.74\text{\hspace{0.17em}N}\end{array}$

Thus, ${\mathrm{T}}_{\mathrm{l}}\text{\hspace{0.17em}is\hspace{0.17em}}8.74\text{\hspace{0.17em}N}$.

The net force in the$\mathrm{y}$ direction is zero. In the$x$-direction, the net force has magnitude as:

${\mathrm{F}}_{\mathrm{net},\mathrm{st}}=({\mathrm{T}}_{\mathrm{u}}+{\mathrm{T}}_{\mathrm{l}})\text{cos}\mathrm{\theta }$

Substitute the values, and we get,

$\begin{array}{l}{\mathrm{F}}_{\mathrm{net},\mathrm{st}}=(35.0\text{\hspace{0.17em}N}+8.74\text{\hspace{0.17em}N})\text{cos}{30.0}^{\text{o}}\\ {\mathrm{F}}_{\mathrm{net},\mathrm{st}}=37.9\text{\hspace{0.17em}N}\end{array}$

Thus, ${\mathrm{F}}_{\mathrm{net},\mathrm{st}}\text{is}37.9\text{\hspace{0.17em}N}$.

The radius of the path is,

$\mathrm{R}=\mathrm{L}\text{cos}\mathrm{\theta }$

Substitute the values, and we get,

$\begin{array}{c}\mathrm{R}=\left(1.70\text{\hspace{0.17em}m}\right)\text{cos}{30}^{\text{o}}\\ =1.47\text{\hspace{0.17em}m}\end{array}$

Using this${\mathrm{F}}_{\mathrm{net},\mathrm{str}}={\mathrm{mv}}^2/\mathrm{R}$, we find the speed of the ball to be,

$\mathrm{v}=\sqrt{\frac{{\mathrm{RF}}_{\mathrm{net},\mathrm{str}}}{\mathrm{m}}}$

Substitute the values, and we get,

$\begin{array}{l}\mathrm{v}=\sqrt{\frac{(1.47\text{m})(37.9\text{N})}{1.34\text{\hspace{0.17em}kg}}}\\ \mathrm{v}=6.4\text{5\hspace{0.17em}m/s}\end{array}$

Thus, $\mathrm{v}\text{\hspace{0.17em}is\hspace{0.17em}}6.45\text{\hspace{0.17em}m/s}$.

The direction of ${\overrightarrow{\mathrm{F}}}_{\mathrm{net},\mathrm{str}}$is leftward (radially inward).