Chapter 5, Problem 12P

To determine

Rank the flywheels in order of the linear speed at the rim, largest to smallest.

Answer to Problem 12P

Since linear speed at the rim is directly proportional to radius/period, Ranking of the flywheels in order of the speed at the rim, from largest to smallest is $\left(c\right)>\left(a\right)=\left(d\right)>\left(b\right)>\left(e\right)$.

Explanation of Solution

Write the expression for the linear speed.

$v=\omega r$ (I)

Here, $v$ is the linear speed, $\omega$ is the angular velocity and $r$ is the radius of spinning object.

Write the expression for the angular velocity.

$\omega =\frac{2\pi }{T}$

Here, $T$ is the time period of rotation.

Substitute $\frac{2\pi }{T}$ for $\omega$ in equation (I) to get $v$.

$v=\frac{2\pi }{T}r$ (II)

Consider option (a)

Radius of flywheel is $8.0\text{cm}$ and period is $4.0\text{ms}$.

Substitute $8.0\text{cm}$ for $r$ and $4.0\text{ms}$ for $T$ in equation(II) to get $v$.

$\begin{array}{c}v=\frac{2\pi }{\left(4.0\text{ms}\times \frac{1\text{s}}{{10}^3\text{ms}}\right)}\text{cm}\left(8.0\text{cm}\times \frac{1\text{m}}{100}\right)\\ =40\pi \text{m}/\text{s}\end{array}$

Therefore, linear velocity flywheel is $40\pi \text{m}/\text{s}$

Consider option (b)

Radius of flywheel is $2.0\text{cm}$ and period is $4.0\text{ms}$.

$v=\frac{2\pi }{T}r$ (II)

Substitute $2.0\text{cm}$ for $r$ and $4.0\text{ms}$ for $T$ in equation(II) to get $v$.

$\begin{array}{c}v=\frac{2\pi }{\left(4.0\text{ms}\times \frac{1\text{s}}{{10}^3\text{ms}}\right)}\text{cm}\left(2.0\text{cm}\times \frac{1\text{m}}{100}\right)\\ =10\pi \text{m}/\text{s}\end{array}$

Therefore, linear velocity flywheel is $10\pi \text{m}/\text{s}$

Consider option (c)

Radius of flywheel is $8.0\text{cm}$ and period is $1.0\text{ms}$.

$v=\frac{2\pi }{T}r$ (II)

Substitute $8.0\text{cm}$ for $r$ and $1.0\text{ms}$ for $T$ in equation(II) to get $v$.

$\begin{array}{c}v=\frac{2\pi }{\left(1.0\text{ms}\times \frac{1\text{s}}{{10}^3\text{ms}}\right)}\text{cm}\left(8.0\text{cm}\times \frac{1\text{m}}{100}\right)\\ =160\pi \text{m}/\text{s}\end{array}$

Therefore, linear velocity flywheel is $160\pi \text{m}/\text{s}$

Consider option (d)

Radius of flywheel is $2.0\text{cm}$ and period is $1.0\text{ms}$.

$v=\frac{2\pi }{T}r$ (II)

Substitute $2.0\text{cm}$ for $r$ and $1.0\text{ms}$ for $T$ in equation(II) to get $v$.

$\begin{array}{c}v=\frac{2\pi }{\left(1.0\text{ms}\times \frac{1\text{s}}{{10}^3\text{ms}}\right)}\text{cm}\left(2.0\text{cm}\times \frac{1\text{m}}{100}\right)\\ =40\pi \text{m}/\text{s}\end{array}$

Therefore, linear velocity flywheel is $40\pi \text{m}/\text{s}$

Consider option (e)

Radius of flywheel is $2.0\text{cm}$ and period is $1.0\text{ms}$.

$v=\frac{2\pi }{T}r$ (II)

Substitute $1.0\text{cm}$ for $r$ and $4.0\text{ms}$ for $T$ in equation(II) to get $v$.

$\begin{array}{c}v=\frac{2\pi }{\left(4.0\text{ms}\times \frac{1\text{s}}{{10}^3\text{ms}}\right)}\text{cm}\left(1.0\text{cm}\times \frac{1\text{m}}{100}\right)\\ =5\pi \text{m}/\text{s}\end{array}$

Therefore, linear velocity flywheel is $5\pi \text{m}/\text{s}$

Conclusion:

The above calculations indicates that flywheel given in option (c)and finally flywheel in option (e)has least velocity.

Since linear speed at the rim is directly proportional to radius/period, Ranking of the flywheels in order of the speed at the rim, from largest to smallest is $\left(c\right)>\left(a\right)=\left(d\right)>\left(b\right)>\left(e\right)$.