Chapter 10, Problem 54A

Answer the previous question for each of the following cases.

a. twice the mass

b. twice the speed

c. twice the length of string (radial distance)

d. twice the mass, twice the speed, and twice the distance all at the same time

To determine

The tension in the string for the given change.

Answer to Problem 54A

The tension in the string is $4.0\text{ N}$ .

Explanation of Solution

Given:

The length of the string is $r=2\text{ m}$ .

The mass is $m=1\text{ kg}$ .

The linear speed is $v=2\text{m}/\text{s}$ .

Also, mass is doubled.

Formula used:

The expression for the tension in the string is,

  $F=\frac{m{v}^2}{r}$

Here, $m$ is the mass, $v$ is the linear speed, and $r$ is the radial distance.

Calculation:

The tension in the string is,

  $\begin{array}{l}F=\frac{m{v}^2}{r}\\ F=\frac{\left(2m\right)v^2}{r}\\ F=\frac{\left(2\times 1\text{kg}\right)\times {\left(\text{2}\text{m}/\text{s}\right)}^2}{2\text{m}}\\ F=4\text{N}\end{array}$

Conclusion:

Thus, the tension in the string is $4.0\text{ N}$ .

To determine

The tension in the string for the given change.

Answer to Problem 54A

The tension in the string is $8.0\text{ N}$ .

Explanation of Solution

Given:

The length of the string is $r=2\text{ m}$ .

The mass is $m=1\text{ kg}$ .

The linear speed is $v=2\text{m}/\text{s}$ .

Speed is doubled.

Formula used:

The expression for the tension in the string is,

  $F=\frac{m{v}^2}{r}$

Here, $m$ is the mass, $v$ is the linear speed, and $r$ is the radial distance.

Calculation:

The tension in the string is,

  $\begin{array}{l}F=\frac{m{v}^2}{r}\\ F=\frac{m{\left(2v\right)}^2}{r}\\ F=\frac{\left(1\text{kg}\right)\times {\left(\text{2}\times \text{2}\text{m}/\text{s}\right)}^2}{2\text{m}}\\ F=8\text{N}\end{array}$

Conclusion:

Thus, the tension in the string is $8.0\text{ N}$ .

To determine

The tension in the string for the given change.

Answer to Problem 54A

The tension in the string is $1.0\text{ N}$ .

Explanation of Solution

Given:

The length of the string is $r=2\text{ m}$ .

The mass is $m=1\text{ kg}$ .

The linear speed is $v=2\text{m}/\text{s}$ .

Length of string or radial distance is doubled.

Formula used:

The expression for the tension in the string is,

  $F=\frac{m{v}^2}{r}$

Here, $m$ is the mass, $v$ is the linear speed, and $r$ is the radial distance.

Calculation:

The tension in the string is,

  $\begin{array}{l}F=\frac{m{v}^2}{r}\\ F=\frac{m{v}^2}{2r}\\ F=\frac{\left(1\text{kg}\right)\times {\left(\text{2}\text{m}/\text{s}\right)}^2}{2\times \left(2\text{m}\right)}\\ F=1\text{N}\end{array}$

Conclusion:

Thus, the tension in the string is $1.0\text{ N}$ .

To determine

The tension in the string for the given change.

Answer to Problem 54A

The tension in the string is $8.0\text{ N}$ .

Explanation of Solution

Given:

The length of the string is $r=2\text{ m}$ .

The mass is $m=1\text{ kg}$ .

The linear speed is $v=2\text{m}/\text{s}$ .

Mass, speed and distance at all the times is doubled.

Formula used:

The expression for the tension in the string is,

  $F=\frac{m{v}^2}{r}$

Here, $m$ is the mass, $v$ is the linear speed, and $r$ is the radial distance.

Calculation:

The tension in the string is,

  $\begin{array}{l}F=\frac{m{v}^2}{r}\\ F=\frac{\left(2m\right){\left(2v\right)}^2}{2r}\\ F=\frac{\left(2\times 1\text{kg}\right)\times {\left(\text{2}\times \text{2}\text{m}/\text{s}\right)}^2}{2\times \left(2\text{m}\right)}\\ F=8\text{N}\end{array}$

Conclusion:

Thus, the tension in the string is $8.0\text{ N}$ .