Angular velocity and angular acceleration
dθ
ds i
CHAPTER 9
!
ri
ROTATION
• Angular velocity and angular acceleration
! equations of rotational motion
• Torque and Moment of Inertia
! Newton’s 2nd Law for rotation
• Determination of the Moment of Inertia
• Rotational kinetic energy
! power
• Rolling objects (with no slip)
θi
The arc length moved by the ith element in a rotating
rigid, non-deformable disk is:
dsi = ri dθ
where dθ is in radians. The angular velocity of the
rotating disk is defined as:
ω=
dθ
,
dt
and so the linear velocity of the i th element (in the
direction of the tangent) is:
vi =
dsi
= riω.
dt
What’s the relationship between the
angular and linear velocity of two points
on the same disk?
DISCUSSION PROBLEM [9.1]:
You have a friend who lives in Minnesota, and you live
in Florida . As the Earth rotates, your linear velocity
is ___________ hers, and your angular velocity is
____________ hers.
A: less than; equal to
RMM02VD2.mov
B: equal to; greater than
C: greater than; less than
The angular velocity ω is the same for all points on the
D: less than; greater than
disk but the linear (or tangential) velocity is not. Look,
E: greater than; equal to
v = rω,
so, since r2 > r1, then v 2 > v1.
If the angular velocity changes
there is angular acceleration ...
dθ
!
a it
!
ai
ds i
!
ri
θi
!
a ir
!
!
a ir ⊥ a it
The angular acceleration of the disk is:
dω d dθ d 2θ
α=
= . =
,
dt dt dt dt 2
and the tangential acceleration of the i th element is:
dv
dω
a it = i = ri
= riα.
dt
dt
Angular velocity ( ω) (vector)
Dimension: ω ⇒
1 [L]
(Check: v i = riω ⇒ [L]
=
).
[T] [T]
Units: rad/s
Angular acceleration ( α) (vector)
Dimension: α ⇒
But, because the ith element is traveling in a circle, it
experiences a radial (centripetal) acceleration:
2
v
a ir (= a ic ) = i = riω 2 .
ri
!
The resultant linear acceleration is a i = a ir 2 + a it 2 .
1
[T]
Units: rad/s2
1
[T]2
CONNECTION BETWEEN LINEAR
AND ROTATIONAL MOTION
Linear motion
Rotational motion
a ⇒ constant
α ⇒ constant
v = v ! + at
ω = ω! + αt
1
(x − x ! ) = v ! t + at 2
2
1
(θ − θ! ) = ω! t + αt 2
2
v 2 = v !2 + 2a(x − x ! )
ω2 = ω!2 + 2α(θ − θ! )
You see, they’re
very similar
Question 9.1: A disk of radius 12 cm, initially at rest,
begins rotating about its axis with a constant angular
acceleration of 3.00 rad/s2 . After 5 s, what are (a) the
angular velocity of the disk, and (b) the tangential and
centripetal accelerations of a point on the perimeter of
the disk? (c) How many revolutions were made by the
disk in those 5 s?
r = 0.12 m: ω! = 0: θ! = 0: t = 5 s
r
α = 3.00 rad/s2 .
ω = ?: θ = ?: a t = ?: a c = ?
In many applications a belt or chain is pulled from or
wound onto a pulley or gear wheel ...
vt
vt
at
(a) ω = ω! + αt = (3.00 rad/s2 )(5 s) = 15.0 rad/s.
(b) v i = riω = (0.12 m)(15.0 rad/s) = 1.80 m/s (linear).
• tangential acceleration: a t = riα
R
= (0.12 m)(3.00 rad/s2 ) = 0.36 m/s 2 .
• centripetal acceleration: a c = riω 2
= (0.12 m)(15.0 rad/s)2 = 27.0 m/s2 .
v2
(Check ... a c =
=
r
m/s)2
(1.80
0.12 m
= 27.0 m/s2 .)
1
(c) (θ − θ! ) = ω! t + αt 2
2
1
= (3.00 rad/s2 )(5 s)2 = 37.5 rad,
2
37.5 rad
⇒n=
= 5.97 rev.
2π
As the string (chain or belt) is removed (or added), its
instantaneous velocity is the same as the tangential
velocity at the rim of the wheel, providing there is no
slip:
i.e., v t = Rω .
Also, under the same conditions, the instantaneous
acceleration of the string is the same as the tangential
acceleration at the rim of the wheel:
dv
dω
i.e., a t = t = R
= Rα.
dt
dt
As we saw in chapter 4 that force produces change in
motion. However, force does not always produce a
[M][L] [M][L]2
Dimension: τ ⇒ [L]
=
[T]2
[T]2
change in rotational motion. It is torque that produces a
(vector)
Unit: N ⋅ m
change in rotational motion. Consider a mass m attached
to a massless rigid rod that rotates around an axis O. The
force F shown will cause the mass to rotate.
Form the radial and tangential components of the
force:
"
F
r
O
"
F
θ
m
ℓ
ℓ = r sin θ
r
O
ℓ
Ft = F sin θ
m
r
O
"
F
θ
m
Fr = F cos θ
ℓ
The magnitude of the torque due to a force F on m is:
τ = ℓF = (r sin θ)F,
Then τ = ℓF = (r sin θ)F = r(F sin θ)
where ℓ is called the lever arm. The lever arm is the
perpendicular distance from the axis of rotation (O) to
the line of action of the force.
i.e., τ = rFt .
Note: the radial component Fr , which passes through
the axis of rotation, does not produce a torque and
"
NOTE: if F passes through O, i.e., ℓ = 0, then τ = 0 and
therefore it does not produce rotation; only the
there will be no change in rotational motion.
in rotation.
tangential component Ft produces a torque that results
Ft = F sin θ
r
O
"
F
θ
m
From above, for each separate mass element, we have
τi = (Δmi )ri2 α,
Fr = F cos θ
ℓ
Newton’s 2nd Law tells us that the tangential component
of the force Ft produces a tangential acceleration a t,
i.e., Ft = ma t .
∴τ = rFt = mra t .
But, from earlier, the tangential (linear) acceleration is
related to the angular acceleration α, viz: a t = rα.
∴τ = mra t = mr 2α.
So torque (τ) produces angular acceleration (α).
ri
O
Δmi
where τi is the net torque
on the ith element.
Summing over all
elements, the total net
torque on the object is:
(
)
τ net = ∑i τi = ∑ i (Δmi )ri2 α = ∑i (Δmi )ri2 α
= Iα,
where we call I = ∑i (Δmi )ri2 the MOMENT OF
INERTIA. This is Newton’s 2nd Law for rotation, i.e.,
τ net = Iα.
A rigid object that rotates about a fixed axis can be
“A net external torque acting on a body produces
thought of as a collection of small, individual elements of
an angular acceleration, α, of that body given
mass ( Δmi ) that each move in a circular path of radius ri,
by Iα, where I is the moment of inertia.”
where ri is measured from the axis of rotation.
(viz: Fnet = ∑i Fi = ma .)
Moment of inertia ... so
what’s that all about?
Dimension: I ⇒ [M][L]2
(scalar)
Units: kg ⋅ m2
Question 9.2: Find the moment of inertia of a uniform
thin rod of length ℓ and mass M rotating about an axis
perpendicular to the rod and through its center.
Every object has a moment of inertia about an axis of
rotation. Its value depends not simply on mass but on
y
how the mass is distributed around that axis. For a
x
discrete collection of n objects, the moment of inertia
about the rotation axis is:
I = ∑n In =
ℓ
∑ n mn rn2 .
For a ‘continuous’ object:
I = Limit ∑ i (Δmi )ri 2 = ∫ r 2dm,
Δmi →0
where m is a function of r.
y
[1] Show for yourselves that the moment of inertia of a
dx
x
− ℓ2
rod of mass M and length ℓ about one end is
x
ℓ
y
2
dx
A rod is a ‘continuous’ object, so the moment of inertia is
2
I = ∫ r dm =
x= ℓ 2
0
2
∫ x dm .
( )
Substituting for dm, the integral becomes
x= ℓ 2
1
I = Mℓ 2 .
3
( )
( )
⎡ ℓ3 (−ℓ3 ) ⎤ 1
2
M
=
ℓ ⎢ 24 − 24 ⎥ = 12 Mℓ .
⎣
⎦
[2] Show for yourselves that the moment of inertia of a
rod of mass M and length ℓ about an axis one-third the
distance from one end is
y
ℓ
⎡ x3 ⎤ 2
I = ∫ x 2 M ℓ dx = M ℓ ⎢ ⎥
⎢⎣ 3 ⎥⎦ − ℓ
x=− ℓ 2
2
( )
ℓ
x= − ℓ 2
The mass per unit length of the rod is M ℓ , so the mass of
the small element of length dx is
dm = M ℓ dx.
x
x
dx
− ℓ3
0
x
1
I = Mℓ 2 .
9
x
2ℓ
3
r
2
The moment of inertia is given by I = ∫ r 2dm. Consider
r1
a ring of radius r and
width dr. If the mass of
dr
Question 9.3: Find the moment of inertia of the circular
r
disk shown below, rotating about an axis perpendicular
the object is M, the
mass of the ring is
2πrdr
dm = M
,
π r22 − r12
to the plane and through its center. The mass of the disk
(
is 1.50 kg.
)
where r1 and r2 are the inner and outer radii of the
object. Then, substituting for dm,
r
2
I = ∫ r 2M
r1
10 cm
2πrdr
2M r2 3
=
∫ r dr
π r22 − r12
r22 − r12 r1
(
) (
)
r
2M ⎡ r 4 ⎤ 2
M
⎢ ⎥ =
= 2
r2 4 − r14
2
2
2
r2 − r1 ⎢⎣ 4 ⎥⎦ r
2 r2 − r1
1
20 cm
(
=
(
2
)
(
)
(
)
M
1
r22 − r12 r22 + r12 = M r22 + r12 .
2
2
− r1
r22
)
(
)(
)
(
)
Sure, but, what’s the significance of I?
1
∴I = M r22 + r12
2
(
)
(
1
= (1.50 kg) (0.20 m)2 + (0.10 m) 2
2
)
Remember, from chapter 4 ...
Mass ⇒ a measure of resistance to a change in
= 3.75 × 10−2 kg ⋅ m2 .
1
[1] If r1 = 0, then Idisk = MR 2 ,
2
linear motion, e.g., how difficult it is to start or
stop linear motion.
R
where R is the radius of the disk ( = r2 ) .
Moment of Inertia ⇒ a measure of resistance to a
change in rotational motion, i.e., how difficult it is
to start or stop rotational motion.
[2] For a thin hoop, r1 ≈ r2 = R,
1
then I hoop = M 2R 2 = MR 2 .
2
( )
R
Providing the thicknesses of the disks are uniform, the
RMM04VD1.MOV
moments of inertia do not depend on thickness. So, these
expressions also apply to cylinders and tubes.
Same torques
Different moments of inertia
Values of the moment of inertia for “simple” shapes ...
Thin rod
Slab
1
I = ML2
12
1
I = M(a 2 + b 2 )
12
L
L
b
b
a
a
Slab
1
I = Ma 2
3
Thin rod
1
I = ML2
3
Question 9.4: A 1 m ruler has a mass of 0.25 kg. A 5 kg
Hollow cylinder
1
2
2
I=
Hollow sphere
2
2
M(R + r )
2
I=
3
MR
Thin-walled cylinder
I = MR 2
Solid cylinder
1
I = MR 2
2
Solid sphere
2
I = MR 2
5
mass is attached to the 100 cm end of the rule. What is
its moment of inertia about the 0 cm end?
O
b
5 kg
DISCUSSION PROBLEM [9.2]:
a
O′
A pair of meter rulers are
The combined moment of inertia is I = I ruler + I mass ,
where I ruler and I mass are calculated about O − O′ .
placed so that their lower
ends are against a wall.
One of the rulers has a
Since a >> b we assume the meter rule is a rod (viz:
1
Question 9.2) with I = Ma 2 about O − O′ .
3
1
1
∴I ruler = Ma 2 = (0.25 kg)(1 m)2
3
3
2
= 0.083 kg ⋅ m .
large mass attached to its
upper end. If the meter
rulers are released at the
same time and allowed to
fall, which one hits the
floor first?
The moment of inertia of the 5 kg mass about O − O′ is
I mass = ma 2 = (5.0 kg)(1 m) 2 = 5.0 kg ⋅ m2 .
Thus, the total moment of inertia of the ruler and mass is
I ruler + I mass = 5.083 kg ⋅ m2 .
A: The meter ruler with the mass.
B: The meter ruler without the mass.
C: They hit the floor at the same time.
Perpendicular axis theorem:
z
Parallel axis theorem:
I
d
Icm
Usually, the moment of inertia is
the center of mass (cm) of the
object. What if the object rotates
about an axis for which we don’t
know the moment of inertia? The
moment of inertia about a general (parallel) axis is given
yi
ri
mi
xi
given for an axis that passes through
Consider a planar object (e.g., a thin
y
disk or sheet) in the x,y plane. By
definition, the moment of inertia
about the z-axis (perpendicular to
x
the plane of the object) is
I z = ∑i mi ri2 = ∑i mi (x i2 + y i2 ) = ∑ i mi x i2 + ∑ i mi y i2
But ∑ i mi x i2 = I x , i.e., the moment of inertia about x, and
∑ i mi y i2 = I y , i.e., the moment of inertia about y.
∴I z = I x + I y .
by:
I = Icm + Md 2
where Icm is the moment of inertia about the center of
mass, M is the mass of the object and d is the distance
between the axes.
Note: the object must be planar
Example of a disk:
z
y
x
Note: the two rotation axes must be parallel
1
I z = MR 2 .
2
But, by symmetry, I x = I y .
1
∴I x = I y = MR 2 .
4
y
y
2m
3 kg
with side length L = 2 m are conncted by massless rods.
4 kg
The masses are m1 = m3 = 3 kg and m2 = m4 = 4 kg.
moment of inertia about an axis that is perpendicular to
the plane of the ensemble and passes through the center
of mass of the system, (c) the moment of inertia about
the x-axis, which passes through m3 and m4.
y
m2
z
cm
2m
D
3 kg
x
z
x
z
(a) Moment of inertia about the z-axis:
I z = ∑i mi ri2 = (3 kg)(2 m)2 + (4 kg)(2 2 m)2
+(3 kg)(2 m)2 + (4 kg)(0) = 56 kg ⋅ m2 .
the square.
∴Icm = ∑i mi ri2 = (3 kg)( 2 m)2 + (4 kg)( 2 m)2
L
m4
2m
(b) By symmetry, the center of mass is at the center of
L
m1
4 kg
2m
Question 9.5: Four masses at the corners of a square
Find (a) the moment of inertia about the z-axis, (b) the
2m
m3
+(3 kg)( 2 m)2 + (4 kg)( 2 m)2 = 28 kg ⋅ m2 .
x
• Check, using the parallel axis-theorem
I z = Icm + MD2
∴Icm = I z − MD 2 = (56 kg ⋅ m2 ) − (14 kg)( 2 m)2
= 28 kg ⋅ m2 .
y
2m
3 kg
4 kg
2m
4 kg
Question 9.6: Four thin rods, each of length ℓ and mass
3 kg
M, are arranged to form a square, in the x,y plane, as
x
shown, with the origin of the axes at the center of the
z
(c) Since the ensemble is planar and confined to the x,y
plane, we can use the perpendicular axis theorem, i.e.,
square. (a) Using the parallel axis theorem, show that
4
I z = Mℓ 2 .
3
(b) Hence find I x and I y.
Iz = Ix + Iy.
But, by symmetry, I x = I y .
(
1
1
∴I x = I z = 56 kg ⋅ m2
2
2
)
z
ℓ
y
2
= 28 kg ⋅ m .
Check:
I x = ∑i mi ri2 = (3 kg )(2 m )2 + (4 kg )(2 m)2
= 28 kg ⋅ m2 .
ℓ
x
We saw in chapter 6 that a linearly moving object has
z
ℓ
ℓ
(a) Using the parallel axis
translational kinetic energy ... an object rotating about
theorem, we have for each
an axis has rotational kinetic energy ...
rod
y
I z = Icm + Md 2 ,
x
where Icm is the moment of
inertia through the center of mass of each rod and d = ℓ 2 .
ℓ2 ⎞
⎛1
4
∴I z (total) = 4⎜ mℓ2 + m ⎟ = mℓ2 .
4⎠ 3
⎝ 12
(b) Since the object is planar we can use the perpendicular
axis theorem, i.e.,
Iz = Ix + Iy.
But because of symmetry I x = I y .
1
2
∴I x = I y = I z = mℓ2 .
2
3
If a rigid object is rotating
ω
O
ri
vi
Δmi
with angular velocity ω,
the kinetic energy of the ith
element is:
1
Ki = (Δmi )v i2
2
1
= (Δmi )(ri2ω 2 ), since v i = riω .
2
So, the total rotational kinetic energy is:
1
K = ∑i Ki = ∑i (Δmi )ri2ω2
2
1
= Iω 2.
2
1
1
* K rot = Iω2 is the analog of K trans = mv 2 .
2
2
A torque is required to rotate (or slow down) an object ...
but torque involves force ...
!
r
!
F
ds
Ft
!
F
γ
θ
dθ
O
Fr
γ + θ = 90"
Ft = F cos γ
Question 9.7: An engine develops 400 N ⋅ m of torque at
3700 rev/min. What is the power developed by the
When force is applied over a distance, work is done,
given by:
! !
dW = F • d s = F.dscos γ = F cos γ.ds
= Ft .ds = Ft .rdθ = τ.dθ
(J or N.m).
Power is the rate at which the torque does work,
dW
dθ
i.e., P =
= τ = τω
(watts).
dt
dt
• dW = τ.dθ is the analog of dW = F.ds.
• P = τω is the analog of P = Fv.
engine?
Torque τ = 400 N ⋅ m.
Angular velocity ω = 3700 rev/min
2π(3700 rev/min)
=
= 387.5 rad/s.
60 s/min
From earlier, power P = τω
5
= (400 N ⋅ m)(387.5 rad/s) = 1.55 × 10 watts.
But 746 watts = 1 HP
∴P =
1.55 ×105 watts
= 208 HP.
746 watts/HP
Question 9.8: An electric motor exerts a constant torque
of 10 N ⋅ m to the shaft of a grindstone. If the moment of
inertia of the grindstone is 2 kg ⋅ m2 and the system starts
from rest, find (a) the rotational kinetic energy of the
grindstone after 8.0 s, (b) the work done by the motor
during this time, and (c) the average power delivered by
the motor.
(a) To find the rotational kinetic energy we need to know
the angular velocity ω.
τ 10 N ⋅ m
Since τ = Iα, α = =
= 5.0 rad/s2 .
2
I 2 kg ⋅ m
The grindstone starts from rest ( ω! = 0) so
ω = αt = (5.0 rad/s2 )(8.0 s) = 40 rad/s.
1
1
∴K = Iω2 = (2 kg ⋅ m2 )(40 rad/s)2 = 1600 J.
2
2
(b) There are two ways to determine the work done by
the motor.
(i) By the work-kinetic energy theorem we would
expect the motor to have done 1600 J of work.
(ii) We can use the expression W = τθ, but we need
to find θ, i.e., the angle through which the grindstone has
turned in 8.0 s.
1
1
θ = ω! t + αt 2 = (5.0 rad/s2 )(8.0 s)2 = 160 rad
2
2
( = 25.46 rev).
∴W = τθ = (10 N ⋅ m)(160 rad) = 1600 J.
(c) The average power is the total work done divided by
the total time interval, i.e.,
ΔW 1600 J
Pav =
=
= 200 W.
Δt
8.0 s
Note: the expression P = τω we derived earlier is actually
the instantaneous power. We cannot use that expression
here as ω is not constant.
Note that the instantaneous power increases linearly from
zero at t = 0 to (10 N ⋅ m) × (40 rad/s) = 400 W at
t = 8.0 s.
2v cm
If a rigid object is suspended from an arbitrary point O
and is free to rotate about that point, it will turn until the
center of mass is vertically beneath the suspension point.
Consider a ball, cylinder, wheel
v cm
ω
v=0
or disc) rolling on a surface
without slipping.
y
• The point in contact with the surface has zero
O
instantaneous velocity relative to the surface.
rcm × cm
xcm
Mg
x
• The velocity of the cm is v cm ( = rω = v),
• The velocity of a point at the top is 2v cm (= 2v).
Since the object has the same linear velocity as the cm,
If the y-direction is vertical and the suspension point is
not at the center of mass, the object will experience a net
i.e., v, the total kinetic energy is:
1
1
K tot = mv 2 + Iω2
2
2
torque given by
τ = Mgx cm,
where x cm is the x component of the center of mass.
Therefore, the object will rotate until x cm = 0, i.e., the
suspension point is direction above the center of mass.
translational + rotational
where ω = v r . So, as a ball rolls down a hill ...
v, ω
Gravitational potential energy ⇒
translational energy
+ rotational energy
Consider rolling a sphere, cylinder and a hoop down an
incline. Do they have the same velocity at the bottom?
h
v
For each object, conservation of energy gives:
1
1
mgh = mv 2 + Iω 2,
2
2
and with no slip v = Rω , where R is the radius of the
object. After manipulation we find:
2gh
v2 =
.
⎛ I
⎞
1+ ⎜
⎟
⎝ mR 2 ⎠
∴ For maximum velocity: I ⇒ smallest value.
RMA07VD2.MOV
2gh
⎛ I
⎞
1+ ⎜
⎟
2
⎝ mR ⎠
2
For a solid sphere I = mR 2 .
5
2gh
10gh
∴v 2 =
,
i.e.,
v
=
.
1 + 25
7
v2 =
Since this is independent of m and R, a bowling ball and
a pool ball would have the same speeds at the bottom of
So, in a race between objects rolling down a slope, the
order would be (1) sphere, (2) cylinder, (3) hoop, and is
completely independent of m and R!
an incline, despite their different masses and radii!
(a) Draw the free body diagram of all the forces acting on
the ball. Use Newton’s 2nd Law down the incline:
Mg sin θ − f = Ma cm ... (i)
N
Question 9.9: A uniform solid ball of mass M and radius R
rolls without slipping down an incline at an angle θ to the
horizontal. Find (a) the frictional force acting at the point
θ
O
f
an expression for a cm, we take
R
torques about O, the center of
Mg
the ball.
of contact with the surface, and (b) the acceleration of the
center of mass of the ball, in terms of M, R, g and θ.
where a cm is the acceleration
of the center of mass. To get
τ = fR = Iα, i.e., α =
fR
,
I
where I is the moment of inertia of the ball and α its
θ
NOTE: there must be friction at the point of contact
otherwise the ball would slide down the incline!
angular acceleration. (NOTE: the normal force N and the
weight force Mg do not contribute to the torque as their
lines of action pass through O.) With no slip
a cm = Rα =
fR 2
I
... ... (ii)
Substituting for a cm in equation (i), we get
Mg sin θ − f = M
fR 2
MfR 2
5
=
= f.
⎛2
⎞
I
⎜ MR 2 ⎟ 2
⎝5
⎠
5
7
∴Mg sin θ = f + f = f ,
2
2
2
i.e., f = Mg sin θ.
7
Note that this is a static frictional force (because there is
no slip).
(b) Substituting for f in equation (ii), we get
fR 2 ⎛ 2
⎞ R2
5
a cm =
= ⎜ Mg sin θ⎟
= gsin θ.
⎛
2
⎞
⎝7
⎠
I
⎜ MR 2 ⎟ 7
⎝5
⎠
You can show yourselves that if we put I = βMR 2 , where
β = 1 for a hoop, β = 12 for a disk, etc., then a general
expression for the static frictional force acting on an
object rolling down an incline, with no slip, and the
acceleration of the center of mass are:
Mg sin θ
gsin θ
f=
and
a
=
.
cm
1+ β
1 + β −1
Question 9.10: Suppose you can choose wheels of any
design for a soapbox derby race car. If the total weight
of the vehicle is fixed, which type of wheel design should
you choose if you want to have the best chance to win
the race?
The total mechanical energy of the car is:
translational kinetic energy + rotational kinetic energy.
Conservation of energy gives:
1
1
Mgh = K trans + K rot = Mv 2 + Iω 2 ,
2
2
F
where M is the total mass of the car and I is the moment of
inertia of the wheels. With four wheels of radius r, for
h
R
example, we have, using the no-slip condition,
1
1
v 2
K rot = (4I)ω 2 = 4 × βmr 2 ⎛ ⎞ = 2βmv 2 ,
⎝ r⎠
2
2
(
)
1
∴Mgh = Mv 2 + 2βmv 2 ,
2
2Mgh
i.e.,
v2 =
.
(M + 4βm)
So, for maximum speed (v) at the bottom of the hill (and
greater average speed) with M fixed, we want m and β to
be as small as possible. Note: that the result does not
depend on the radius of the wheels (r). So, solid wheels
(β = 12 ) are a good choice with the mass concentrated as
close to the axle as possible.
Question 9.11: A cue ball is struck by a horizontal cue a
distance h above the center of the ball. If the cue ball is
to roll without slipping, what is h? Express your answer
in terms of the radius R of the ball.
You can assume that the frictional force of the table on
the ball is negligible compared with the applied force F.
F
h
R
•
f
The net torque about the center is τ = Fh − fR.
From earlier, if there is no-slip then v cm = Rω,
dv
dω
i.e., a cm = cm = R
= Rα.
dt
dt
Using Newton’s 2nd Law F + f = ma cm.
If F >> f , then F ≈ ma cm and τ ≈ Fh = Iα.
F
Fh
I
∴ = a cm = Rα = R , i.e., h =
.
m
I
mR
2
For a solid sphere I = mR 2 .
5
2
∴h = R.
5
2
• h > R ⇒ top spin
5
“SLIP”
2
• h < R ⇒ back spin
5
m2 30 kg
20 kg m1
2m
Question 9.12: The 30 kg mass, shown above, is
released from rest, from a distance of 2 m above the
ground. Modeling the pulley as a uniform disk with a
radius of 10 cm and mass 5 kg, find (a) the speed of the
30 kg mass just before it strikes the ground, (b) the
angular velocity of the pulley at that instant, (c) the
tensions in the two strings, and (d) the time it takes for
the 30 kg block to reach the ground. Assume the
bearings in the pulley are frictionless and there is no slip
between the string and the pulley.
•
•
m2 30 kg
20 kg m1
2m
v
ω
•
m1
m2 30 kg
m2
v
We have both translational and rotational motion. The
v
2m
20 kg m1
m1
m2
v
continued ...
i.e., 589 = 10v 2 + 392 +15v 2 + 1.3v 2
moment of inertia of the pulley is:
1
1
I = mR 2 = (5 kg)(0.10 m) 2 = 0.025 kg ⋅ m2 .
2
2
(a) Using conservation of energy, as m2 hits the ground
1
1
1
m2 gh = m1v 2 + m1gh + m2 v 2 + Iω 2 .
2
2
2
v
The angular velocity of the pulley ω = , so
R
1
(30 kg)(9.81 m/s 2 )(2 m) = (20 kg)v 2
2
1
+(20 kg)(9.81 m/s2 )(2 m) + (30 kg)v 2
2
⎛
⎞
1
v2
⎟
+ (0.025 kg ⋅ m2 )⎜⎜
2⎟
2
⎝ (0.10 m) ⎠
ω
•
∴v =
(b) ω =
(c)
197
= 2.74 m/s.
26.3
v 2.74
=
= 27.4 rad/s.
R 0.10
T1
m1g
T2
a
m2 g
a
Draw the free-body
diagrams for the masses.
What is the upward acceleration of m1?
v 2 = 2ah.
v 2 (2.74 m/s)2
∴a =
=
= 1.88 m/s2 .
2h
2(2 m)
T1
m1g
Then T1 − m1g = m1a,
T2
a
m2 g
i.e., T1 = m1 (g + a) = 234 N.
a
Also T2 − m2 g = m2 (−a),
i.e., T2 = m2 (g − a) = 238 N.
1
(d) ( y − y ! ) = h = at 2 .
2
∴t =
2h
2(2 m)
=
= 1.46 s.
a
1.88 m/s2