Chapter 1: Q22 E (page 296)

You are to design a rotating cylindrical axle to lift 800-N buckets of cement from the ground to a rooftop 78.0 m above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise.
(a) What should the diameter of the axle be to raise the buckets at a steady 2.00 cm/s when it is turning at 7.5 rpm?
(b)If instead the axle must give the buckets an upward acceleration of 0.400 m/s², what should the angular acceleration of the axle be?

Short Answer

Expert verified

The diameter of the axle is $5.09 c m$ .

Thus, the Angular acceleration of the axle is $15.7 \frac{r a d}{s^{2}}$ .

Step by step solution

Step:-1 explanation

In the given question $v = 2.00 \frac{c m}{s}$ .

$ω = 7.5 r e v$

We know that the formula,

$v = r ω$ _______(1)

Step:-2 Concept

The diameter of the axle is to raise the buckets at a steady 2.00 cm/s.

We use here equation (1)

v=r $ω$

Put the value v and $ω$ in the expression.

2.00=Rrole="math" localid="1667971386345" $(\frac{7.5 r e v}{\min}) (\frac{1 m i n}{60 s}) (\frac{2 π r a d}{1 r e v})$

r=2.55cm

We know that thed=2r.

Step:-3 calculation

Here we get the value of the D,

d=2 $\times$ 2.55

=5.09cm

The diameter of the axle is 5.09cm.

Step:-4(b) explanation

The angular acceleration of the axle.

We know that the angular formula is,

$a_{t a n} = r_{α}$

Given $a_{\tan} = 0.400 r = 0.0255$

Step:-5(b) Concept

The angular acceleration of the axle.

We know the angular acceleration formula $α = \frac{a_{\tan}}{r}$ .

Step:-6(b) calculation

Put the value in the above expression,

$= \frac{0.400 \frac{m}{s^{2}}}{0.0255 m} = 15.7 \frac{r a d}{s^{2}}$

Thus, Angular acceleration of the axle is $15.7 \frac{r a d}{s^{2}}$ .

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Short Answer

Step by step solution

Step:-1 explanation

Step:-2 Concept

Step:-3 calculation

Step:-4(b) explanation

Step:-5(b) Concept

Step:-6(b) calculation

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