Q5-58E - University Physics with Modern Physics

Question

A bowling ball weighing 71.2 N (16.0 lb) is attached to the ceiling by a 3.80-m rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is $4.20\text{\hspace{0.17em}m/s}$. At this instant, what are 

(a) the acceleration of the bowling ball, in magnitude and direction, and 

(b) the tension in the rope?

Step-by-Step Solution

Answer

(a) The acceleration of the bowling ball is $4.642{\text{\hspace{0.17em}m/s}}^{\text{2}}$, in an upward direction.

(b) The tension in the rope is, $104.926\text{\hspace{0.17em}N}$.   

Step 1: Identification of the given data

 The given data can be listed below as,

• The weight of the bowling ball is, $m=71.2\text{\hspace{0.17em}N}$.
• The radius of the rope is, $R=3.80\text{\hspace{0.17em}m}$.
• The speed of the bowling ball is, $v=4.20\text{\hspace{0.17em}m/s}$.

Step 2: Significance of tension

It is the pulling force transmitted axially or transmitted through any wire, cable, string rope, etc. when it is pulling at the end is known as Tension force

Step 3: Determination of the acceleration of the bowling ball

(a)

 

The expression for the acceleration when the particle moves in a circular path can be expressed as,

 $a_{rad}=\frac{v^2}{R}$

Here $R$ is the radius of the object in a circular path $v$, is the speed of the object.

 

Substitute $4.20\text{\hspace{0.17em}m/s}$ for $v$ , and  $3.80\text{\hspace{0.17em}m}$ for $R$ in the above equation.

$\begin{array}{c}{a}_{rad}=\frac{{\left(4.20\text{\hspace{0.17em}m/s}\right)}^2}{3.8\text{\hspace{0.17em}m}}\\ =4.642{\text{\hspace{0.17em}m/s}}^{\text{2}}\end{array}$

Hence, the required acceleration is,$4.642{\text{\hspace{0.17em}m/s}}^{\text{2}}$ in an upward direction.

Step 4: Determination of the tension in the rope

(b)

 

The expression for Newton’s second law using tension can be expressed as,

$\begin{array}{c}\sum F=ma\\ T-mg=ma\\ T=m(a+g)\\ T=\frac{W}{g}(a+g)\end{array}$

Here is the acceleration,$g$ is the acceleration due to gravity, and $W$ is the weight of the ball.

 

Substitute $4.642{\text{\hspace{0.17em}m/s}}^{\text{2}}$ for $a$, $9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$ for $g$, and $71.2\text{\hspace{0.17em}N}$ for $W$ in the above equation.

 $\begin{array}{c}T=\frac{71.2\text{\hspace{0.17em}N}}{9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}}\times (4.642{\text{\hspace{0.17em}m/s}}^{\text{2}}+9.8{\text{\hspace{0.17em}m/s}}^{\text{2}})\\ =104.926\text{\hspace{0.17em}N}\end{array}$

Hence, the required tension is, $104.926\text{\hspace{0.17em}N}$.