HP 48GII - Angular Mechanics (4, 2); Centripetal Force (4, 3); Hookes Law (4, 4)

Equation Reference 5-23
Angular Mechanics (4, 2)
Equations:
Example:
Given: I=1750_lb
∗ in^2, Θ =360_°, r=3.5_in, α=10.5_r/min^2, ω i=0_r / s.
Solution: r=1.1017E–3_ft_
∗ lbf, Ki=0_ft ∗ lbf, W=6.9221E–3_ft ∗ lbf, Kf=6.9221E–3_ft ∗ lbf,
at=8.5069E–4_ft/s^2, Ni=0_rpm,
ω f=11.4868_r/min, t=1.0940_min, Nf=1.8282_rpm, Pavg=1.9174E–7_hp.
Centripetal Force (4, 3)
Equations:
Example:
Given: m=1_kg, r=5_cm, N=2000_Hz.
Solution:
ω =12566.3706_r/s, ar=7895683.5209_m/s, F=7895683.5209_N, v=628.3185_m/s.
Hooke's Law (4, 4)
The force is that exerted by the spring.
Equations:
Fk–x⋅=
W
1–
2
------
kx
2
⋅⋅=
Example:
Given: k=1725_lbf/in, x=125_in.
Solution: F=-2156.25_lbf, W=-112.3047_ft
∗ lbf.
τ
I
α ⋅
= K i
1
2
- - -
I
ω
i
2
⋅ ⋅
=
K
f
1
2
- - -
I
ω
f
2
⋅ ⋅
=
W K
f
K i
–
=
ω
f
ω
i
α
+ t
⋅
=
P
τ ω
⋅
=
P a v g
W
t
-----
=
W
r
Θ⋅
=
a
t
α
r
⋅
=
ω
f
2
π
N
f
⋅ ⋅
=
ω
2
π
N
⋅ ⋅
=
ω
i 2
π
N
i
⋅
⋅
=
F m
ω
2
r
⋅ ⋅
=
ω
2
π
N
⋅ ⋅
=
ω
v
r
- - -
=
a r
v
2
r
-----
=