The weight of the disk is 30 lb and its k
G
equals 0.6 ft.
The spring has a stiffness of 2 lb/ft
and an unstretched
length of 1 ft. Find the velocity at the instant G moves 3
ft to the left.
The disk is released from rest in the
position shown and rolls without slipping.
(1) Initial Position
(2) Final
Position
g1
+ V
e1
+ U
12
= T
2
+ V
+ V
e2
What equation, and what terms are zero?
No change in height, so Vg is zero.
Remember, rolling friction doesn’t do any
work (point at bottom doesn’t displace).
If
you have slip, then the F=
μ
k
N DOES do
work.
Set up your Ve terms.
V
e1
= ½ k (s
1
)
2
= ½ (2)(51)
2
= 16 J
V
e2
= ½ 2 (3)
2
= 9 J
T
2
= ½ m (v
G2
)
2
+ ½ I
G
(
ϖ
2
)
2
= ½ (30/32.2) (v
G2
)
2
+ ½ (30/32.2) 0.6
2
(
ϖ
2
)
2
What is T
2
?
CG has a velocity
Two unknowns – how can I relate?
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/22/2011 for the course ME 212 taught by Professor Staff during the Spring '05 term at Cal Poly.
 Spring '05
 staff

Click to edit the document details