This preview shows page 1. Sign up to view the full content.
which yields 2
v
= +44 m/s. This is consistent with Fig. 113(c).
(i) We can proceed as in part (h) or simply recall that the bottommost point is in firm
contact with the (zerovelocity) road. Either way – the answer is zero.
(j) The translational motion of the center is constant; it does not accelerate.
(k) Since we are transforming between constantvelocity frames of reference, the
accelerations are unaffected. The answer is as it was in part (e): 1.5
×
10
3
m/s
2
.
(1) As explained in part (k),
a
= 1.5
×
10
3
m/s
2
.
1. The velocity of the car is a constant
()
ˆˆ
80 km/h (1000 m/km)(1 h/3600 s) i
( 22m s)i,
v
=+
= +
G
and the radius of the wheel is
r
= 0.66/2 = 0.33 m.
(a) In the car’s reference frame (where the lady perceives herself to be at rest) the road is
moving towards the rear at
G
vv
road
ms
=− =−
22
, and the motion of the tire is purely
rotational. In this frame, the center of the tire is “fixed” so
v
center
= 0.
(b) Since the tire’s motion is only rotational (not translational) in this frame, Eq. 1018
gives
top
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

Click to edit the document details