4.)! A
hanging mass is attached to a string which is
threaded over a massive pulley and attached to a
second mass sitting on a frictionless incline. Your
finger keeps everything stationary. Known is:!
1
m1 , m h , m p , R, g, θ, and Icm of pulley = m p R 2
2
a.) Ignoring the forces acting at the pulley s
pin, draw a f.b.d. identifying all the forces acting
on both masses and the pulley.!
1
mpR 2
2
e.) The hanging mass drops from rest a distance h.
What is its velocity magnitude by the end of the drop?!
m1 , m h , m p , R, g, θ, and Icm of pulley =
mp
Ffinger
m1
mh
θ
mp
m1
θ
mh
h
f.) What is the angular velocity of the pulley at that point?!
b.) Determine the finger force required to just hold the system in equilibrium.!
g.) What is the angular momentum of the pulley at that point?!
1.)!
1
mpR 2
2
c.) The finger is removed and the system begins to
accelerate. What is its acceleration magnitude? (Assume
the acceleration is down the incline.)!
m1 , m h , m p , R, g, θ, and Icm of pulley =
mp
m1
θ
mh
d.) What is the pulley s angular acceleration?!
2.)!
3.)!
m1 , m h , m p , R, g, θ, and Icm of pulley =
4.)! A
hanging mass is attached to a string which is
mp
threaded over a massive pulley and attached to a
second mass sitting on a frictionless incline. Your
Ffinger
m1
finger keeps everything stationary. Known is:!
mh
1
θ
m1 , m h , m p , R, g, θ, and Icm of pulley = m p R 2
2
a.) Ignoring the forces acting at the pulley s pin, draw a f.b.d. identifying all the forces acting
on both masses and the pulley.!
T1
T2
T1
m1
T2
pulley
∑Γ
pulley
− T2 R + T1R = Ipinα
T1
m hg
b.) Determine the finger force required to just hold the system in equilibrium.!
Using the quick and dirty approach for N.S.L., the only forces motivating the system to
accelerate are the finger, the component of gravity acting on the tabled mass and gravity
on the hanging mass. Assuming the finger force is positive and the acceleration is equal to
0
zero, we get:!
⇒
⎛1
⎞⎛ a⎞
− T2 R + T1R = ⎜ m p R 2 ⎟ ⎜ ⎟
⎝2
⎠ ⎝ R⎠
⇒
− T2 + T1 =
⇒
− ( m hg + m h a ) + ( m1g sin θ − m1a ) =
⇒ a=
Ffinger − m1g sin θ + m hg = ( m1 + m h ) a
1
mpa
2
−m hg + m1g sin θ
⎛1
⎞
⎜⎝ m p + m1 + m h ⎟⎠
2
d.) What is the pulley s angular acceleration? !
⇒ Ffinger = m1g sin θ − m hg
mh
θ
T2
N
mp
c.) (con t.)!
Ffinger
m1g
1
mpR 2
2
⎛ a⎞
α=⎜ ⎟
⎝ R⎠
1
mpa
2
(good enough!)!
7.)!
5.)!
1
mpR 2
2
c.) The finger is removed and the system begins to
accelerate. What is its acceleration magnitude?!
m1 , m h , m p , R, g, θ, and Icm of pulley =
mp
m1
θ
m1
mh
Note that if the hanging mass drops, the mass
on the incline goes UP the incline:!
(Assume the acceleration is
down the incline.)!
T1
∑ KE ∑ U
m2
m1g
∑F
1
N
T2
0
⇒
m hg
m1 ,x
T1 − m1g sin θ = −m1a
⇒ T1 = m1g sin θ − m1a
1
mpR 2
2
e.) The hanging mass drops a distance h. What is its
velocity magnitude at that point?!
m1 , m h , m p , R, g, θ, and Icm of pulley =
∑F
T2 − m hg = m h a
⇒ T2 = m hg + m h a
6.)!
+ ∑ Wext =
+ [ m hgh ] +
[ m gh ] =
h
⇒
m h ,y
1
0
∑ KE
mp
m1
mh
θ
2
+
∑U
h
2
1
1
⎡1
⎤
= ⎢ m1v2 + m h v2 + Ipulleyω 2 ⎥ + ⎡⎣ m1g ( h sin θ ) ⎤⎦
2
2
⎣2
⎦
2
⎡1
⎤
1
1⎛1
2
2
2⎞ ⎛ v ⎞
⎢ m1v + m h v + ⎜ m p R ⎟ ⎜ ⎟ ⎥ + ⎡⎣ m1g ( h sin θ ) ⎤⎦
⎝
⎠
⎝
⎠
2
2 2
R ⎥⎦
⎢⎣ 2
v=
2 ( m hgh − m1g ( h sin θ ))
m
m1 + m h + p
2
8.)!
m1 , m h , m p , R, g, θ, and Icm of pulley =
f.) What is the angular velocity of the pulley at that
point?!
ω=
mp
1
mpR 2
2
m1
θ
mh
h
v
R
g.) What is the angular momentum of the pulley at that point?!
L = Ipinω
9.)!