3/10/2014
Torque (Latin for “twist”):
τ = Fr sin θ
To twist an object,
one needs to apply force
r
θ
τ = Fr sin θ
τ = ( F sin θ )r = F⊥ r
τ = F (r sin θ ) = Fr⊥
line of action
The ease of twisting will
depend on
• force
• distance between the
force and the axis
• direction of the force
axis
F
Torque: two perspectives
r⊥ =r sinθ
θ
lever arm
axis
axis
r
r
F⊥ =F sinθ
θ
θ
θ
F
Torque:
Torque and work
θ2
θ
axis
s
r
s
W = ( F sin φr )
r
W = τθ
W = ( F sin φ )r
r
r1
s
Units:
• N-m
φ
F sinφ
φ
θ1
F
F1
P=
torque negative
Example
Lever arm
for F2
∆t
= τω
F2=30 N
30°
F3=20 N
30°
10°
X
45°
F3=20 N
X
τ∆θ
Example continued:
Example: Calculate the torque due to the three forces shown about the left end of
the bar (the red X). The length of the bar is 4m and F2 acts in the middle of the
bar.
F2=30 N
Note analogy with W=Fs
Proof for simple case of one particle
W = ( F sin φ ) s
axis
r2
Multiple forces:
• τnet = τ1+ τ2 + …
W = τθ
torque positive
F2
Sign convention:
• CW (clockwise): negative
• CCW(counter-clockwise):
positive
F
10°
45°
F1=25 N
F1=25 N
Lever arm
for F3
The lever arms are:
5
r1 = 0
r2 = (2m )sin 60° = 1.73 m
r3 = (4m ) sin 10° = 0.695 m
6
1
3/10/2014
Mechanical Energy of Rigid Body
Example continued:
The torques are:
τ1 = 0
τ 2 = +(1.73 m )(30 N ) = +51.9 Nm
τ 3 = +(0.695 m )(20 N ) = +13.9 Nm
The mechanical energy of a rigid body with fixed
axis:
Emec = U + K rot
Gravitational Potential Energy
U = ∑ mi gyi = g ∑ mi yi = gyCM M
The net torque is +65.8 Nm and is the sum of the above results.
Conservation of Energy
Work-Energy
7
Exam 2 Fall 2011: Problem 13
• A thin stick with mass M, length L, and moment
of inertia ML2/3 is hinged at its lower end and
allowed to fall freely as shown in the figure. If
its length L = 2 m and it starts from rest at an
angle θ = 20o, what is the speed (in m/s) of the
free end of the stick when it hits the table?
Answer: 7.43 m/s
% Right: 14%
Ei = MgyCM
L
= Mg cos θ
2
E f = 12 Iω = 12 I
2
f
v
Torque and angular acceleration
τ = Iα
hinge
θ
L
cos θ
2
h=
2
f
2
L
Proof for simple case of one particle
r
Ei = E f
( F sin φ ) r = m(α r )r = ( mr 2 ) α
φ
τ = Iα
F
PHY 2053
Page 9
Static Equilibrium
Center of Gravity (CG)
x
r
If g is the same for all parts of the body then the
CG coincides with the CM.
Fnet,x = ∑ Fx = 0
x1
The gravity force on a body effectively acts at a
single point, called center of gravity (CG)
y
Conditions for static equilibrium:
• Two equations for the forces
• One equation for the torques FN
τ net = ∑ τ = 0
F sin φ = m(α r )
F sinφ
φ
= 3(9.8m / s 2 )(2m) cos(20o ) ≈ 7.43m / s
Fnet,y = ∑ Fy = 0
F sin φ = mat
axis
mgL3 cos θ
mgL3 cosθ
vf =
=
= 3 gL cos θ
2
1
I
3 mL
University of Florida
Note analogy with F=ma
x2
m2g
m1g
FN − m1g − m 2 g = 0
m1gx1 − m 2 gx 2 = 0
Lever
arm
Lever
arm
2
3/10/2014
Equilibrium problem
Example: a ladder against a wall
What must the friction coefficient with the
floor be so that the ladder does not fall
down?
∑F = 0
∑F = 0
∑τ = 0
f s − FN 2 = 0
x
FN1
mg
FN1 − mg = 0
y
z
FN2
θ
L
FN 2 L sin θ − mg cosθ = 0
2
f s = µsFN1 = µsmg
L
µ s mgL sin θ − mg cos θ = 0
2
∑F = 0
∑F = 0
∑τ = 0
x
fs
1
1
µs sin θ − cos θ = 0 µs =
2tan θ
2
Torque and angular acceleration
τ = Iα
A crate of mass M is hanging by a rope from
horizontal beam with mass m. The beam is
supported by a cable attached at an angle θ.
Find the tension in the cable and the force with
which the wall acts on the beam.
Note analogy with F=ma
x:
T cos θ − Fh = 0
T sinθ + Fv − Mg − mg = 0
L
τ:
MgL + mg − TL sin θ = 0
2
(M + 0.5m)g
(M + 0.5m)g
T=
Fh = T cos θ =
sinθ
tanθ
y
F sin φ = mat
r
F sin φ = m(α r )
( F sin φ ) r = m(α r )r = ( mr 2 ) α
F sinφ
φ
φ
F
τ = Iα
T
Fh
θ
Mg
mg
M
Fv = (M + m)g − (M + 0.5m)g = 0.5mg
Example: Bicycle wheel
Example (text problem 8.53): A bicycle wheel (a hoop) of radius 0.3 m and mass 2
kg is rotating at 4.00 rev/sec. After 50 sec the wheel comes to a stop because of
friction. What is the magnitude of the average torque due to frictional forces?
Proof for simple case of one particle
axis
Fv
y:
∑τ = Iα = MR α
2
α=
ωi = 4.00
rev  2π rad 

 = 25.1 rad/sec
sec  1 rev 
ωf = 0
∆ω ω f − ωi
=
= −0.50 rad/s 2
∆t
∆t
τ av = MR 2 α = 0.09 Nm
16
3