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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 5.09cm.

Thus, the Angular acceleration of the axle is15.7rads2.

Step by step solution

01

Step:-1 explanation

In the given question v=2.00cms.

蝇=7.5rev

We know that the formula,

v=r蝇_______(1)

02

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蝇

Put the value v and蝇in the expression.

2.00=Rrole="math" localid="1667971386345" 7.5revmin1min60s2蟺rad1rev

r=2.55cm

We know that thed=2r.

03

Step:-3 calculation 

Here we get the value of the D,

d=2脳2.55

=5.09cm

The diameter of the axle is 5.09cm.

04

Step:-4(b) explanation 

The angular acceleration of the axle.

We know that the angular formula is,

atan=r伪

Given atan=0.400r=0.0255

05

Step:-5(b) Concept  

The angular acceleration of the axle.

We know the angular acceleration formula伪=atanr.

06

Step:-6(b) calculation 

Put the value in the above expression,

=0.400ms20.0255m=15.7rads2

Thus, Angular acceleration of the axle is 15.7rads2.

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