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Trigonometry Speeds of pulley and a belt

karush

Well-known member
Jan 31, 2012
2,714
A pulley has a radius of $12.96\text { cm}$
it takes $18\text { s}$ for $56\text { cm}$ of belt to go around the pulley.

(a) find the angular speed of the pulley in $\displaystyle\frac{\text {rad}}{\text{s}}$

well from $\displaystyle\frac{56\text { cm}}{18\text{ s}}
\approx \frac{3.11\text { cm}}{\text{s}}$

and $\displaystyle\text {rad}=\frac{S}{r}
=\frac{3.11\text { cm}}{12.96\text { cm}}

\approx 0.34 \text{ rad}$

since $\text{S}$ is the arc length for one $\text{s}$ then

$\displaystyle\approx \frac{0.34\text {rad}}{\text{s}}$ angular speed

(b) find the linear speed of the belt in $\displaystyle\frac{\text {cm}}{\text{s}}
\approx \frac{3.11\text { cm}}{\text{s}}$

well if correct?? it seem a little bit choppy way to solve it.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
A pulley has a radius of $12.96\text { cm}$
it takes $18\text { s}$ for $56\text { cm}$ of belt to go around the pulley.

(a) find the angular speed of the pulley in $\displaystyle\frac{\text {rad}}{\text{s}}$

well from $\displaystyle\frac{56\text { cm}}{18\text{ s}}
\approx \frac{3.11\text { cm}}{\text{s}}$

and $\displaystyle\text {rad}=\frac{S}{r}
=\frac{3.11\text { cm}}{12.96\text { cm}}

\approx 0.34 \text{ rad}$

since $\text{S}$ is the arc length for one $\text{s}$ then

$\displaystyle\approx \frac{0.34\text {rad}}{\text{s}}$ angular speed

(b) find the linear speed of the belt in $\displaystyle\frac{\text {cm}}{\text{s}}
\approx \frac{3.11\text { cm}}{\text{s}}$

well if correct?? it seem a little bit choppy way to solve it.
Yeah. It's a bit choppy. ;)

There is a difference between quantities and units.
Speed is a quantity, rad is a unit.
Note that rad is a rather special unit, since it's a dimensionless unit like "rev".
Take care though, since you have $2\pi \text{ rad}$ in one $\text{rev}$.

The usual symbol for the quantity angular speed is $\omega$, which has the unit $\text{rad/s}$.

Btw, can I assume that with S you meant the speed of the belt?
I'm asking since the usual symbol for speed is v.

Anyway, in that case you have for (a):
$$\omega = \frac S r = \frac {3.11\frac{\text{cm}}{\text{s}}}{12.96\text { cm}} = 0.24 \frac{\text{rad}}{\text{s}}$$
So I'm afraid your answer is not quite right, neither numerically, nor in the specification of the units.

Your answer for (b) is correct though.
 

karush

Well-known member
Jan 31, 2012
2,714
Btw, can I assume that with S you meant the speed of the belt?
I'm asking since the usual symbol for speed is v.
OK, well, from the book i am looking at
$\text{s} = \text{arc length}$ so $\displaystyle\text{a}=\frac{s}{r}$

where $\text{r}=$ radius
and $\text{a}=$ angle in degrees or radians

this is confusing since $\text{s}$ looks like it is speed or seconds but is arc length
which I capitalized earlier to distinguish from speed or seconds.
so what is meant is.


$\displaystyle\text {rad}=\frac{s}{r} =\frac{3.11\text { cm}}{12.96\text { cm}} \approx 0.34 \text{ rad}$

and so

$\displaystyle\omega = \frac{a}{t} = \frac {0.34\text{rad}}{sec}$

or is this fog over choppy waters
the notation is kinda well..
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
ok, well, from the book i am looking at
$\text{s} = \text{arc length}$ so $\displaystyle\text{a}=\frac{s}{r}$

where $\text{r}=$ radius
and $\text{a}=$ angle in degrees or radians

this is confusing since $\text{s}$ looks like it is speed or seconds but is arc length
which I capitalized earlier to distinguish from speed or seconds.
so what is meant is.


$\displaystyle\text {rad}=\frac{s}{r} =\frac{3.11\text { cm}}{12.96\text { cm}} \approx 0.34 \text{ rad}$

and so

$\displaystyle\omega = \frac{a}{t} = \frac {0.34\text{rad}}{sec}$

or is this fog over choppy waters
the notation is kinda well..
Ah okay.
Actually you can distinguish the distance $s$ from seconds $\text{s}$ by italic versus upright.
It's unusual though to use $a$ as an angle. The symbol $a$ is usually an acceleration. Angle is usually denoted as $\phi$ or $\theta$.

Anyway, what you should have is that in a period of $t=1\text{ s}$ the belt travels a distance $s=3.11\text { cm}$ along a radius $r=12.96\text { cm}$.
The corresponding angle is:
$$a =\frac{s}{r} =\frac{3.11\text { cm}}{12.96\text { cm}} = 0.24 \text{ rad}$$
Therefore the corresponding angular speed is:
$$\omega = \frac{a}{t} = 0.24 \frac {\text{rad}}{\text{s}}$$


Alternatively you could say that in a period of $t=18\text{ s}$ the belt travels a distance $s=56\text{ cm}$ along a radius of $r=12.96\text { cm}$.
Then the corresponding angle is:
$$a =\frac{s}{r} =\frac{56\text { cm}}{12.96\text { cm}} = \frac{56}{12.96} \text{ rad}$$
And the corresponding angular speed is:
$$\omega = \frac{a}{t} = \frac {\frac{56}{12.96}}{18} \frac{\text{rad}}{\text{s}} = 0.24 \frac {\text{rad}}{\text{s}}$$
 

karush

Well-known member
Jan 31, 2012
2,714
OK, think i am getting the picture... so .24 not .34 calc error

you were a great help these text books are sometimes shy on info to understand....

I post a couple more of these to make sure I have it down...