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?

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

role="math" localid="1664352126983" $\omega =7.5rev$

We know that the formula,

$v=r\omega$_______(1)

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

We use here equation (1)

$v=r\omega$

Put the value v and$\omega$ in the expression.

$$\begin{gathered} 2.00=R\left(\frac{7.5rev}{min}\right)\left(\frac{1min}{60s}\right)\left(\frac{2\mathrm{\pi rad}}{1rev}\right) \\ r=2.55cm \end{gathered}$$

We know that the d=2r.

Here we get the value of the D,

$$\begin{gathered} d=2\times 2.55 \\ =5.09cm \end{gathered}$$

The diameter of the axle is 5.09cm.

The angular acceleration of the axle.

We know that the angular formula is,

$a_{\tan }=r_{\alpha }$

Given$a_{\tan }=0.400$

$r=0.0255$

The angular acceleration of the axle.

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

Put the value in the above expression,

$$\begin{gathered} =\frac{0.400{\displaystyle \frac{m}{s^2}}}{0.0255m} \\ =15.7\frac{rad}{s^2} \end{gathered}$$

Thus, Angular acceleration of the axle is data-custom-editor="chemistry" $=15.7\frac{rad}{s^2}$.