This preview shows page 1. Sign up to view the full content.
101. (a) The distance traveled by the coin in 3.14 s is 3(2
π
r
) = 6
(0.050) = 0.94 m. Thus,
its speed is
v
= 0.94/3.14 = 0.30 m/s.
(b) The centripetal acceleration is given by Eq. 617:
22
2
(0.30 m/s)
1.8 m/s .
0.050 m
v
a
r
==
=
(c) The acceleration vector (at any instant) is horizontal and points from the coin towards
the center of the turntable.
(d) The only horizontal force acting on the coin is static friction
f
s
and must be large
enough to supply the acceleration of part (b) for the
m
= 0.0020 kg coin. Using Newton’s
second law,
()
23
0.0020 kg 1.8 m/s
3.6
10
N
s
fm
a
−
=×
.
(e) The static friction
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Acceleration

Click to edit the document details