Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

Solutions 121

in contact with the bobbin all the time, the
velocity of point C relative to the board is di-
rected along the board, whence wR tan-1 (a/2) =
vo sin a. Since there is no slipping of the bobbin
over the horizontal surface, we can write

vo

R RI-r •

Therefore, we obtain the following expression for
the angular velocity ce:

v a 2v sin 2 (a/2)

vi = sin a • tan

R

+r

2 = (R- 1 - r) cos (a/2)

1.17. The area of the spool occupied by the wound

thick tape is S1 = n (r 2 — r) = 81[r?. Then the

length of the wound tape is 1 = = 8a (r1/d),

where d is the thickness of the thick tape.

The area of the spool occupied by the wound

thin tape is S2 = n (ri 2 — r?), where ri is the final

radius of the wound part in the latter case. Since
the lengths of the tapes are equal, and the tape
thickness in the latter case is half that in the
former case, we can write

1=

27( (r' 2 -1)

rf

'2 2 z^2

Consequently, the final radius ri of the wound part

in the latter case is

1/5 r1.

The numbers of turns N, and N2 of the spool
for the former and latter winding can be written as

2r1 ( ri

, N2 —

wheuce is = (^) 1) ti.