Phys 221 SI :
Torque
Leader:
Course:
Instructor:
Date:
Supplemental Instruction
Iowa State University
Michael
Phys 221
Prell
03/05/2012
Useful Equations
⃗
⃗ NET (Equivalent to Newton's Second Law) ; ⃗τ =⃗r × F
✔ Torque : ⃗τ NET =I α
✔ “” Without Slipping : x=r θ ; v=r ω ; a=r α
✔ Common Moments of Inertia :
1
2
◦ Solid Cylinder I = M R ; Hollow Cylinder I =M R2
2
2
1
2
2
◦ Solid Sphere I = M R ; Rod (about center) I = M L
5
12
1
2
◦ Rod (about end) I = M L
3
Problems :
1. (*) Four masses are set rolling down an incline, rank them according to the order in
which they arrive at the bottom :
• #1) A solid cylinder of mass m and radius r.
1
1
2
2
I= M R = mr
•
2
2
• #2) A solid sphere of mass 2m and radius r/2.
2
2
r 2 1
•
I = M R 2= ( 2 m )
= mr 2
5
5
2
5
• #3) A hollow cylinder of mass 2m and radius r.
I =M R2=( 2 m )( r )2=2 m r 2
•
• #4) A solid sphere of mass m and radius 2r.
2
8
I = ( m )( 2 r )2= m r 2
•
5
5
• A larger moment of inertia (angular mass) will result in a smaller angular acceleration
(if subjected to the same torque), therefore the one which will reach the bottom fastest
has the smallest moment of inertia
I 2<I 1<I 4<I 3 → 3 → 4 → 1 → 2
•
()
2. (*) We have a pulley of mass mp and radius rp over which we string a massless rope. We
intend to use the rope to lift a mass m at a constant speed. What is the minimum force
that we need to pull the rope with?
• Newton's 2nd Law on mass m : T 1−m g =m a NET =0
• Torque on pulley : τ NET = I α=0=r ( F −T 1 ) (B/c it is at constant angular velocity)
• Therefore, F =T 1=m g
Supplemental Instruction
1060 Hixson-Lied Student Success Center v 294-6624 v www.si.iastate.edu
3. (**) A solid sphere of mass 18 kg and radius 1 cm rolls without slipping on an incline of
30o which is 10 meters long. If the sphere starts from rest, what is the acceleration of the
sphere when it is halfway down the incline? What is the force due to friction?
• Forces are : F⃗N =〈 0, F N 〉 , F⃗ f =〈− f s , 0〉 , F⃗g =〈 m g sin θ ,−m g cos θ〉
⃗ NET =〈 m g sin θ− f s , F N −m g cos θ〉=〈 ma NET , 0〉
F
•
• Torques are : τ⃗N =0 , τ⃗f = f s r , τ⃗g =0
⃗τ NET = f s r =I α
•
• If we take the moment of inertia to be proportional to a constant I =k m r 2 and
assume that the sphere is not slipping a NET =r α , then this expression becomes
a
f
f s r =( k m r 2 ) NET =k m r a NET → a NET = s
r
km
• Plugging this into the expression above, m a NET =m g sin θ− f s →
fs
m g sin θ
1
m
=m g sin θ− f s → f s 1+ =m g sin θ → f s =
k
1
km
1+
k
f s m g sin θ 1
1
• And a NET =
=
=
g sin θ
km
1 k m k +1
1+
k
2
• For the case of a solid sphere, k =
→
5
1
1
5
2
a NET =
g sin θ=
g sin θ= g sin θ and f s = m g sin θ
2/ 5+1
7/5
7
7
5
2
o
2
• With numbers : a NET = (9.8 m/ s ) sin(30 )≈3.5 m/ s
7
2
2
o
f s = (18 kg )(9.8 m/ s )sin (30 )≈25.2 N
•
7
( )
( )
( )
4. (***) A 5.0 kg block hangs at the end of a massless ideal rope wound around a uniform
cylinder with mass 6.0 kg and radius 20 cm. The cylinder can only rotate about its axis.
What is the tension in the cable when the system is released?
• Forces : m g−T =m a NET (only one-dimension)
• Torques : τ=r T =I α
1
2
• For a uniform cylinder, I = M r and because it is rolling w/o slipping,
2
a NET =α r
a
1
1
•
r T = M r 2 NET → T = M a NET
2
2
r
m g −T
m g −T
T
=2
• Combining this with the expression above, a NET =
→
m
m
M
mM
g
m M g −M T =2 mT → T ( M +2 m ) =m M g → T =
•
M +2 m
( 5.0 kg )(6.0 kg)
• Plugging in numbers, T =
(9.8 m/ s 2)≈18.4 N
(6.0 kg )+2(5.0 kg)
( )