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Trigonometry Linear speed of a flywheel

karush

Well-known member
Jan 31, 2012
2,770
A flywheel with a $15cm$ diameter is rotating at a rate of $\displaystyle\frac{7 rad}{s}$
What is the linear speed of a point on the rim, in $\displaystyle\frac{cm}{min} $.

$s=r\theta$ so $7.5(7) = 152$cm
then $\displaystyle v=\frac{s}{t}=\frac{152cm}{s}\cdot\frac{60s}{min}=\frac{1320cm}{min}$

I am not sure just what a Radian (rad) is in this, so hope I didn't make this to simple. don't have answer so hope mine ok
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Your method is correct (but you have made some arithmetical errors)...I would write:

\(\displaystyle v=r\omega=\frac{15}{2}\text{ cm}\cdot7\frac{1}{\text{s}}\cdot\frac{60\text{ s}}{1\text{ min}}=?\)
 

karush

Well-known member
Jan 31, 2012
2,770
Your method is correct (but you have made some arithmetical errors)...I would write:

\(\displaystyle v=r\omega=\frac{15}{2}\text{ cm}\cdot7\frac{1}{\text{s}}\cdot\frac{60\text{ s}}{1\text{ min}}=?\)

$\displaystyle\frac{3150 cm}{min}$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$\displaystyle\frac{3150 cm}{min}$
Correct. The method you used is:

\(\displaystyle v=\frac{s}{t}=\frac{r\theta}{t}=r\frac{\theta}{t}\)

Now defining the angular velocity $\omega$ to be:

\(\displaystyle \omega=\frac{\theta}{t}\)

we then have:

\(\displaystyle v=r\omega\)

That is, the linear velocity $v$ is the product of the radius of motion and the angular velocity.

Did you find the error in your previous calculations?
 

karush

Well-known member
Jan 31, 2012
2,770
let me see if this set up ok

a wheel with $30cm$ radius is rotating at a rate of $\displaystyle{3rad}{s}$ what is v in $\displaystyle\frac{m}{s}$


$\displaystyle v=r\omega$
$\displaystyle 30\text{ cm}\cdot3\frac{1}{\text{s}}\cdot \frac{m}{100\text{cm}}=$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
let me see if this set up ok

a wheel with $30cm$ radius is rotating at a rate of $\displaystyle{3rad}{s}$ what is v in $\displaystyle\frac{m}{s}$


$\displaystyle v=r\omega$
$\displaystyle 30\text{ cm}\cdot3\frac{1}{\text{s}}\cdot \frac{m}{100\text{cm}}=$
Yes, that is correct. (Clapping)
 

karush

Well-known member
Jan 31, 2012
2,770
oops just noticed the ans should be in \(\displaystyle \frac{\text {m}}{\text {min}}\)

so...

$\displaystyle 30\text{ cm}\cdot \frac{3}{\text{s}}
\cdot \frac{60 \text { s}}{\text { min}}
\cdot \frac{\text { m}}{100\text{ cm}}=\frac{54 \text {m}}{\text {min}}$

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Did you find the error in your previous calculations?
yes I had 152 cm it should be 52.5 cm