Section 10.6: Torque
Translation-Rotation Analogues & Connections
Translation
Rotation
Displacement
x
θ
Velocity
v
ω
Acceleration
a
α
Mass (moment of inertia) m
I
Kinetic Energy (K) (½)mv2
(½)Iω2
Force
F
?
CONNECTIONS
s = rθ, v = rω, at= rα, ac = (v2/r) = ω2r
• Newton’s 1st Law (rotational language version): “A
rotating body will continue to rotate at a
constant angular velocity unless an external
TORQUE acts.”
• Clearly, to understand this, we need to
define the concept of TORQUE.
• Newton’s 2nd Law (rotational language version):
Also needs torque.
• To cause an object to rotate requires a FORCE,
F. (Cause of angular acceleration α).
• BUT: The location of the force on the object &
the direction it acts are also important!

Introduce the torque concept.
 From experiment!
• Angular acceleration α  F.

• But also α  (distance from the point of application
of F to the hinge = Moment Arm or Lever Arm, d)
• Moment Arm  d = distance of the axis of
rotation from the “line of action” of force F
• d = Distance which is  to both the axis of rotation
and to an imaginary line drawn along the direction of
the force (“Line of Action”).
• Line of Action  Imaginary line extending out both ends of
the force vector. Experiment finds that angular acceleration
α  (force) (moment arm) = Fd

Define: TORQUE
Lower case
Greek “tau” 
τ  Fd
τ causes α
(Just as in the linear motion case, F causes a)
Newton’s laws & rotational motion
We want to find a rotational analogue to force
The figure is a top view of a
door that is hinged on the left:
The 4 pushing forces are of
equal strength. Which of these
will be the most effective at
opening the door?




F1 will open the door.
F2 will not.
F3 will open the door, but not as easily as F1.
F4 will open the door – it has same magnitude as F1, but
we know it is not as effective as pushing at the outer
edge of the door.
Ability of force F to cause a rotation or twisting
motion depends on 3 factors:
1. The magnitude F of the force.
2. The distance r from point of
application to pivot.
3. The angle at which F is applied.
Make these idea quantitative.
Figure is a force F applied at
one point on a rigid body.

We define a new quantity called
torque, denoted by the symbol  :
  rF sin 
where  is the angle from r to F .
τ depends on the 3 properties, & is the
rotational analogy to force.
Let’s see this again
  rF sin 
Torque Units:
Newton-meter = N m
Sign convention: A torque that tends to rotate an object in a counterclockwise
direction is positive. torque that tends to rotate an object in clockwise direction
is positive. See Figure.
Section 10.7: Rigid Object Under a Net Torque
Consider the object shown.
• The force F1 will tend to cause
a counterclockwise rotation about O.
• The force F2 will tend to cause
a clockwise rotation about O
• The net torque is:
τnet = ∑τ = τ1+ τ2
= F1d1 – F2d2
We’ve seen that toque is the rotational analogue to force.
Now we need to learn what toque does.
Figure is a model rocket engine attached to one end of lightweight, rigid rod.
Tangential acceleration at & the
angular acceleration α are related.
at  r
Ft  mat  mr
Tangential component of force.
Multiply both sides by r:
rFt  mr 2
   mr  
  rFt
2
Therefore, torque τ causes angular acceleration α. This
relation is analogous to Newton’s 2nd Law F = ma
Newton’s 2nd Law for Rotations
Figure: A rigid body undergoes pure
rotational motion about a fixed, frictionless,
& unmoving axis. Net torque on the object is
the sum of the torques on all the individual
particles.
 net

2
  i    mi ri      mi ri  
i
i
 i

2
Definition: Moment of Inertia
I  m1r12  m2 r2 2  m3r32 
  mi ri 2
i

 net
I
or  net  I
Newton’s 2nd law for rotational motion
Ex. 10.7: Net Torque on a Cylinder
Double cylinder, shaped as shown.
Attached to axle through center.
Large part, radius R1 has rope around
it & is pulled, with tension T1 to right.
Small part, radius R2 has a rope
around it & is pulled down with
tension T2.
(A) Net torque?
∑τ = τ1+ τ2 = T2R2 – T1R1
(B) T1 = 5 N, to R1 = 1 m, T2 = 15 N, R2 = 0.5 m.
Net torque? Which direction will it rotate?
∑τ = (15)(0.5) – (1)(5) = 2.5 N m
Direction is counterclockwise (positive torque)
Ex. 10.8: Rotating Rod
Ex. 10.10: Angular Acceleration of a Wheel
Wheel, radius R, mass M, moment of inertia I. A cord is
wrapped around it & attached to mass m. System is
released & m falls & wheel rotates. Find the tension T in
the cord, acceleration a of falling m, angular acceleration
α of wheel.
Newton’s 2nd Law for wheel:
∑τ = Iα (1)
Tension T is tangential force on wheel & is the only force
producing a torque. So:
∑τ = TR (2)
m moves in a straight line
Newton’s 2nd Law for m: SFy = ma = mg – T (3)
No slipping of cord.

a = αR
(4)
From (1) & (2):
α = (∑τ/I) = (TR)/I (5)
From (4) & (5):
a = (mg – T)/m = (TR2)/I
(6)
Solve for T from (6):
T = (mg)/[1 + (mR2/I)]
Put into (6) to get a:
a = g/[1 + (I/mR2)]
Put into (4) to get α:
α = g/[R + (I/mR)]
a