MHB Calculating Angular and Linear Speeds of Pulley and Belt

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The discussion focuses on calculating the angular and linear speeds of a pulley with a radius of 12.96 cm, where 56 cm of belt travels around it in 18 seconds. The initial calculation for angular speed yielded approximately 0.34 rad/s, but further clarification revealed the correct angular speed is 0.24 rad/s. The linear speed of the belt was confirmed to be approximately 3.11 cm/s. Participants emphasized the importance of distinguishing between arc length and speed in calculations, as well as the correct notation for angular measurements. Overall, the conversation highlights common misunderstandings in physics notation and calculations.
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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.
 
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karush said:
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.
 
I like Serena said:
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:
karush said:
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:
$$\omega = \frac{a}{t} = \frac {\frac{56}{12.96}}{18} \frac{\text{rad}}{\text{s}} = 0.24 \frac {\text{rad}}{\text{s}}$$
 
OK, think i am getting the picture... so .24 not .34 calc error

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

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