Chapter 16, Problem 78A

(a)

To determine

The tension in the string when the object is at the bottom of its path.

Answer to Problem 78A

  $T=2.1\times {10}^2\text{ N}$

Explanation of Solution

Given:

Mass of an object is $m=2.0\text{ kg}$

Length of a string to which an object is attached is $l\text{ = 1}\text{.5 m}$

Speed of a swinging object is $v=\text{ 12 m/s}$

Formula used:

The tension in the string when the object is at the bottom of its path can be calculated by,

  ${\text{T = F}}_{cp}+F_g$

  $\left(1\right)$

Where,

${\text{F}}_{cp}$ is the centripetal force acting on a revolving object, ${\text{F}}_{cp}=\frac{m{v}^2}{r}$ , and

$F_g$ is the force exerted by the Earth’s gravity on an object, $F_g=mg$

Substituting for ${\text{F}}_{cp}$ and $F_g$ in equation $\left(1\right)$ ,

  $T=\frac{m{v}^2}{r}+mg=m\left[\frac{v^2}{r}+g\right]$

  $\left(2\right)$

Where,

$m$ is mass of an object,

$v$ is the speed of revolution of an object

$r$ is the radius of the circle in which an object in revolving, In this situation it is $l\text{ = 1}\text{.5 m}$

$g$ is acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

Calculation:

Substituting the numerical values in equation $\left(2\right)$ ,

  $T=\left(2.0\text{ kg}\right)\left[\frac{{\left(\text{12 m/s}\right)}^2}{\text{1}\text{.5 m}}+9.8{\text{ m/s}}^2\right]$

  $=\left(2.0\text{ kg}\right)\left[96{\text{ m/s}}^2+9.8{\text{ m/s}}^2\right]$

  $=\left(2.0\text{ kg}\right)\left(105.8{\text{ m/s}}^2\right)$

  $=211.6\text{ kg}{\text{.m/s}}^2$

  $=2.1\times {10}^2\text{ N}$

Conclusion:

The tension in the string when the object is at the bottom of its path is $2.1\times {10}^2\text{ N}$ .

(b)

To determine

The tension in the string when the object is at the top of its path.

Answer to Problem 78A

  $T=1.72\times {10}^2\text{ N}$

Explanation of Solution

Given:

Mass of an object is $m=2.0\text{ kg}$

Length of a string to which an object is attached is $l\text{ = 1}\text{.5 m}$

Speed of a swinging object is $v=\text{ 12 m/s}$

Formula used:

The tension in the string when the object is at the bottom of its path can be calculated by,

  ${\text{T = F}}_{cp}-F_g$

  $\left(3\right)$

Where,

${\text{F}}_{cp}$ is the centripetal force acting on a revolving object, ${\text{F}}_{cp}=\frac{m{v}^2}{r}$ , and

$F_g$ is the force exerted by the Earth’s gravity on an object, $F_g=mg$

Substituting for ${\text{F}}_{cp}$ and $F_g$ in equation $\left(3\right)$ ,

  $T=\frac{m{v}^2}{r}-mg=m\left[\frac{v^2}{r}-g\right]$

  $\left(4\right)$

Where,

$m$ is mass of an object,

$v$ is the speed of revolution of an object

$r$ is the radius of the circle in which an object in revolving, In this situation it is $l\text{ = 1}\text{.5 m}$

$g$ is acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

Calculation:

Substituting the numerical values in equation $\left(4\right)$ ,

  $T=\left(2.0\text{ kg}\right)\left[\frac{{\left(\text{12 m/s}\right)}^2}{\text{1}\text{.5 m}}-9.8{\text{ m/s}}^2\right]$

  $=\left(2.0\text{ kg}\right)\left[96{\text{ m/s}}^2-9.8{\text{ m/s}}^2\right]$

  $=\left(2.0\text{ kg}\right)\left(\text{86}{\text{.2 m/s}}^2\right)$

  $=172.4\text{ kg}{\text{.m/s}}^2$

  $=1.72\times {10}^2\text{ N}$

Conclusion:

The tension in the string when the object is at the top of its path is $1.72\times {10}^2\text{ N}$ .