Transcript File - Phy 2048-0002

Rotational Kinematics
and Energy
Rotational Motion
Up until now we have been looking at the
kinematics and dynamics of translational motion –
that is, motion without rotation. Now we will widen
our view of the natural world to include objects that
both rotate and translate.
We will develop descriptions (equations) that
describe rotational motion
Now we can look at motion of bicycle wheels,
roundabouts and divers.
I.
Rotational variables
- Angular position, displacement, velocity, acceleration
II. Rotation with constant angular acceleration
III. Relation between linear and angular variables
- Position, speed, acceleration
IV. Kinetic energy of rotation
V. Rotational inertia
VI. Torque
VII. Newton’s second law for rotation
VIII. Work and rotational kinetic energy
Rotational kinematics
• In the kinematics of rotation we encounter
new kinematic quantities
– Angular displacement
q
– Angular speed
w
– Angular acceleration
a
– Rotational Inertia
I
– Torque
t
• All these quantities are defined relative to
an axis of rotation
Angular displacement
• Measured in radians or degrees
• There is no dimension
Dq = qf - qi
qi
Dq
Axis of rotation
2
Dq  q f  q i 
rad
3
CW
qf
Angular displacement and arc length
• Arc length depends
on the distance it is
measured away
from the axis of
rotation
2
Dq  q f  q i 
r
3
S p  Dq  rp
r
S q  Dq  rq
s
Dq 
r
qi
P
Q
sp
sq
Axis of rotation
qf
CW
Angular Speed
•
Angular speed is the rate of change of
angular position
Dq
w
Dt
•
We can also define the instantaneous
angular speed
Dq
w  lim
Dt 0 Dt
Average angular velocity and
tangential speed
• Recall that speed is distance divided by time
elapsed
• Tangential speed is arc length divided by time
elapsed
s
vt 
Dt
s
• And because Dq  we can write
r
Dq  s 1 vT
w 


Dt
r Dt r
Average Angular Acceleration
• Rate of change of angular velocity
Dw w f  w i
a

Dt
Dt
• Instantaneous angular acceleration
Dw
a  lim
Dt  0 D t
Angular acceleration and tangential
acceleration
• We can find a link between tangential
acceleration at and angular acceleration α
vf
vi

w

w
Dw
aT
f
i
r
r
a



Dt
Dt
Dt
r
• So
aT
a
r
Centripetal acceleration
• We have that
vT
ac 
r
• But we also know that
• So we can also say
2
v w r
2
T
ac  w r
2
2 2
Example: Rotation
• A dryer rotates at 120 rpm. What distance do your
clothes travel during one half hour of drying time in
a 70 cm diameter dryer? What angle is swept out?
Example: Rotation
• A dryer rotates at 120 rpm. What distance do your
clothes travel during one half hour of drying time in
a 70 cm diameter dryer? What angle is swept out?
• Angle: Dq wDt = 120 r/min x 0.5 x 60 min =
120x2(rad) /min x 60 min/h x 0.5 h = 2.3 x 104 r
– Distance: s = Dq r and w = Dq/Dt so s = wDtr
– s = 120x2(rad)/min x 60 min/h x 0.5 h x 0.35
m
= 1.3 km
Rotational motion with constant
angular acceleration
• We will consider cases where a is constant
• Definitions of rotational and translational
quantities look similar
• The kinematic equations describing rotational
motion also look similar
• Each of the translational kinematic equations
has a rotational analogue
Rotational and Translational
Kinematic Equations
v f  v i  at
Dx  v i t  at
1
2
2
v f  v  2aDx
2
2
i
vi  vf
v
2
Constant a motion
What is the angular acceleration of a car’s wheels
(radius 25 cm) when a car accelerates from 2 m/s to 5
m/s in 8 seconds?
Constant a motion
What is the angular acceleration of a car’s wheels (radius
25 cm) when a car accelerates from 2 m/s to 5 m/s in 8
seconds?
vi 2m / s
wi  
 8rad / s
r 0.25m
v f 5m / s
wf  
 20rad / s
r 0.25m
w f  wi (20  8)rad / s
a

 1.5rad / s 2
Dt
8s
A rotating wheel requires 3.00 s to rotate through
37.0 revolutions. Its angular speed at the end of the
3.00-s interval is 98.0 rad/s. What is the constant
angular acceleration of the wheel?
A rotating wheel requires 3.00 s to rotate through
37.0 revolutions. Its angular speed at the end of the
3.00-s interval is 98.0 rad/s. What is the constant
angular acceleration of the wheel?
1 2
q f  qi  wit a t
2
(
w f  wi  a t
wi  w f  a t

1 2
1 2
q f  qi  w f  at t at  w ft at
2
2
1
 2 rad 
2
37.0 rev 
 98.0 rad s( 3.00 s  a ( 3.00 s



1 rev
(
2

232 rad  294 rad  4.50 s2 a
61.5 rad
2
a

13.
7
r
ad
s
4.50 s2
Example: Centripetal Acceleration
• A 1000 kg car goes around a bend that has a radius of
100 m, travelling at 50 km/h. What is the centripetal
force? What keeps the car on the bend? [What keeps
the skater in the arc?]
Example: Centripetal Acceleration
• A 1000 kg car goes around a bend that has a radius of
100 m, travelling at 50 km/h. What is the centripetal
force? What keeps the car on the bend? [What keeps
the skater in the arc?]
2
 km 1000 
1000 kg   50


2
mv
h 3600 

Fc 

 1929 N
r
100 m
Friction keeps the car and skater on the bend
Car rounding a bend
Frictional force of road on tires supplies
centripetal force
If ms between road and tires is lowered then
frictional force may not be enough to provide
centripetal force…car will skid
Locking wheels makes things worse as
mk < ms
Banking of roads at corners reduces the risk
of skidding…
Car rounding a bend
Horizontal component of the normal force
of the road on the car can provide the
centripetal force
Ncosq
If
N
2
mv
N sin q 
r
then no friction is required
Nsinq
q
Fg
An automobile makes a turn whose radius is 150 m. The
road is banked at an angle of 18°, and the coefficient of
friction between the wheels and road is 0.3. Find the
maximum speed for the car to stay on the road without
skidding the banked road.
An automobile makes a turn whose radius is 150 m. The
road is banked at an angle of 18°, and the coefficient of
friction between the wheels and road is 0.3. Find the
maximum speed for the car to stay on the road without
skidding the banked road.
x-component:
FN sin q + fs cos q = ma = mv2/R;
y-component: FN cos q – fs sin q – mg = 0.
FN cos q – msFN sin q = mg,
FN = mg/(cos q – ms sin q).
An automobile makes a turn whose radius is 150 m. The
road is banked at an angle of 18°, and the coefficient of
friction between the wheels and road is 0.3. Find the
maximum speed for the car to stay on the road without
skidding the banked road.
From the x-equation we get:
g (sin q  m s cos q )

R
cos q  m s sin q
2
vmax
2
vmax

(150m)(9.8m / s 2 )(sin 180  0.3 cos 180 )
vmax  32m / s
(cos 180  0.3 sin 180 )
Rotational Dynamics
• Easier to move door at A than at B using the
same force F
hinge
A
B
• More torque is exerted at A than at B
Torque
• Torque is the rotational analogue of Force
• Torque, t, is defined to be
t = Fr
Where F is the force applied tangent to the
rotation and r is the distance from the axis of
rotation
F
r
Torque
• A general definition of torque is
F
q
t = Fsinq r
r
• Units of torque are Nm
• Sign convention used with torque
– Torque is positive if object tends to rotate CCW
– Torque is negative if object tends to rotate CW
Condition for Equilibrium
• We know that if an object is in (translational)
equilibrium then it does not accelerate. We
can say that SF = 0
• An object in rotational equilibrium does not
change its rotational speed. In this case we
can say that there is no net torque or in other
words that:
St = 0
Torque and angular acceleration
• An unbalanced torque (t) gives rise to an
angular acceleration (a)
• We can find an expression analogous to
F = ma that relates t and a
• We can see that
Ft = mat
• and Ftr = matr = mr2a
Ft
(since at = ra
r m
• Therefore
t = mr2a
Torque and Angular Acceleration
t = mr2a
Angular acceleration is directly proportional to the net
torque, but the constant of proportionality has to do with
both the mass of the object and the distance of the object
from the axis of rotation – in this case the constant is mr2
This constant is called the moment of inertia. Its
symbol is I, and its units are kgm2
I depends on the arrangement of the rotating system. It
might be different when the same mass is rotating about a
different axis
Newton’s Second Law for Rotation
• We now have that
t = Ia
• Where I is a constant related to the
distribution of mass in the rotating
system
• This is a new version of Newton’s
second law that applies to rotation
A fish takes a line and pulls it with a tension of 15 N for 20
seconds. The spool has a radius of 7.5 cm. If the moment
of inertia of the reel is 10 kgm2, through how many rotations
does the reel spin? (Assume there is no friction)
A fish takes a line and pulls it with a tension of 15 N for 20
seconds. The spool has a radius of 7.5 cm. If the moment
of inertia of the reel is 10 kgm2, through how many rotations
does the reel spin? (Assume there is no friction)
The torque and corresponding angular acceleration are
t  Fr  15 N  0.075m  1.125 N  m
t 1.125 Nm
rad 2
a 

0
.
1125
2
I
s
10kgm
Using rotational kinematics, the angle through which the
spool spins is
If we assume the spool starts from rest (ωi= 0) then
q  0 . 5a t  0 . 5  0 . 1125 s  20  22 . 5 rad
2
2
2
which is 3.5 rotations (divide 22.5 by 2π rotation)
Angular Acceleration and I
The angular acceleration reached by a
rotating object depends on M, r, (their
distribution) and T
T
When objects are rolling under the influence
of gravity, only the mass distribution and the
radius are important
Moments of Inertia for Rotating Objects
I for a small mass m rotating about a point a
distance r away is mr2
What is the moment of inertia for an object that
is rotating –such as a rolling object?
Disc?
Sphere?
Hoop?
Cylinder?
Moments of Inertia for Rotating
Objects
The total torque on a rotating system is the sum
of the torques acting on all particles of the
system about the axis of rotation –
and since a is the same for all particles:
I  Smr2 = m1r12+ m2r22+ m3r32+…
Axis of rotation
Continuous Objects
To calculate the moment of inertia for
continuous objects, we imagine the object to
consist of a continuum of very small mass
elements dm. Thus the finite sum Σmi r2i
becomes the integral
I   r dm
2
Moment of Inertia of a Uniform Rod
Lets find the moment of inertia of a uniform
rod of length L and mass M about an axis
perpendicular to the rod and through one end.
Assume that the rod has negligible thickness.
L
Moment of Inertia of a Uniform Rod
We choose a mass element dm at a distance x
from the axes. The mass per unit length (linear
mass density) is λ = M / L
Moment of Inertia of a Uniform Rod
dm = λ dx
M
dm 
dx
L
L
L
L
M
M 2
I   x dm   x
dx 
x dx

L
L 0
0
0
2
M 1

x
L 3
2
3 L
0
3
M L
1
2

 ML
L 3
3
Example:Moment of Inertia of a Dumbbell
A dumbbell consist of point masses 2kg and 1kg attached by a
rigid massless rod of length 0.6m. Calculate the rotational inertia
of the dumbbell (a) about the axis going through the center of
the mass and (b) going through the 2kg mass.
Example:Moment of Inertia of a Dumbbell
m1 x1  m2 x2 (2kg)(0)  (1kg)(0.6m)
X 

 0.2m
m1  m2
2kg  1kg
(a)
I CM  m1( x1  X CM )2  m2 ( x2  X CM )2
 (2kg)(0m  0.2m)2  (1kg)(0.6m  0.2m)2  0.24kgm2
Example:Moment of Inertia of a Dumbbell
(b)
I  m2 L2  (1kg)(0.6m) 2  0.4kgm2
Moment of Inertia of a Uniform Hoop
dm
All mass of the hoop M
is at distance r = R from
the axis
R

M
I  r dm 
2

0
M
R dm  R
2
2

0
dm  R
2
M
m0
 MR
2
Moment of Inertia of a Uniform Disc
We expect that I will
be smaller than MR2
since the mass is
uniformly distributed
from r = 0 to r = R
rather than being
concentrated at r = R
as it is in the hoop.
dr
r
R
Each mass element is a hoop of radius r and
thickness dr. Mass per unit area
σ = M / A = M /πR2
Moment of Inertia of a Uniform Disc
M
dm  dA 
2rdr
A
R
dr
R
I   r dm   r  2rdr  2  r dr
2
2
3
0
0
r
R
2M R
M
1
4
2
I 

R  MR
2
A 4
2R
2
4
Moments of inertia I for Different Mass Arrangements
Moments of inertia I for Different Mass Arrangements
Which one will win?
A hoop, disc and sphere are all rolled
down an inclined plane. Which one will win?
Which one will win?
A hoop, disc and sphere are all rolled
down an inclined plane. Which one will win?
1. Hoop I = MR2
2. Disc
I = ½MR2
3. Sphere I = 2/5MR2
a = t/ I
1. a1 = t/ MR2
2. α2= 2(t/ MR2)
3.a3 = 2.5(t/ MR2)
I. Rotational variables
Rigid body: body that can rotate with all its parts locked
together and without shape changes.
Rotation axis: every point of a body moves in a circle
whose center lies on the rotation axis. Every point moves
through the same angle during a particular time interval.
Reference line: fixed in the body, perpendicular to the
rotation axis and rotating with the body.
Angular position: the angle of the reference line relative
to the positive direction of the x-axis.
q
arc length
radius
s

r
Units: radians (rad)
2r

1 rev  360 
 2 rad
r

1 rad  57.3  0.159 rev
Note: we do not reset θ to
zero with each complete
rotation of the reference line
about the rotation axis. 2
turns  θ =4π
Translation: body’s movement described by x(t).
Rotation: body’s movement given by θ(t) = angular position
of the body’s reference line as function of time.
Angular displacement: body’s rotation about its axis
changing the angular position from θ1 to θ2.
D q  q 2  q1
Clockwise rotation  negative
Counterclockwise rotation  positive
Angular velocity:
Average:
w avg
Instantaneous:
q 2  q1 D q


t 2  t1
Dt
Dq dq
w  lim

Dt  0 D t
dt
Units: rad/s or
rev/s
These equations hold not only for the rotating rigid body
as a whole but also for every particle of that body
because they are all locked together.
Angular speed (ω): magnitude of the angular velocity.
Angular acceleration:
Average:
Instantaneous:
a avg
w 2  w1 D w


t 2  t1
Dt
Dw dw
a  lim

Dt  0 D t
dt
Angular quantities are “normally” vector
quantities  right hand rule.
Examples: angular velocity,
angular acceleration
Exception: angular displacements
Object rotates around the direction of
the vector  a vector defines an axis
of rotation not the direction in which
something is moving.
II. Rotation with constant angular
acceleration
Linear equations
Angular equations
v  v0  at
w  w0  at
1 2
x  x0  v0t  at
2
2
2
v  v0  2a( x  x0 )
1 2
q  q 0  w0 t  at
2
w 2  w02  2a (q  q 0 )
1
x  x0  (v0  v)t
2
1 2
x  x0  vt  at
2
1
q  q 0  (w0  w )t
2
1 2
q  q 0  wt  at
2
III. Relation between linear and angular variables
Position:
Speed:
s q r
θ always in radians
ds
dq
r
 v  w r
dt
dt
ω in rad/s
v is tangent to the circle in which a point
moves
Since all points within a rigid body have the
same angular speed ω, points with greater
radius have greater linear speed, v.
If ω=cte  v=cte  each point within the body undergoes
uniform circular motion
Period of revolution:
T
2 r
v
2

w
Acceleration:
dv d w

r  a  r  at  a  r
dt
dt
Tangential component
of linear acceleration
Radial component of
linear acceleration:
v 2 w 2r 2
2
ar 

w r
r
r
Units: m/s2
Responsible for changes in the direction of the linear
velocity vector v
Kinetic energy of rotation
Reminder: Angular velocity, ω
particles within the rotating body.
is the same for all
Linear velocity, v of a particle within the rigid body
depends on the particle’s distance to the rotation axis (r).
1
2 1
2 1
K  m1v1  m2v2  m3v32  ... 
2
2
2

1
1
1

mi vi2 
mi (w  ri ) 2  
2
2
2
i
i




i

mi ri2 w 2


Moment of Inertia
Kinetic energy of a body in pure rotation
1 2
K  Iw
2
Kinetic energy of a body in pure translation
1
2
K  MvCOM
2
Discrete rigid body
 I =∑miri2
Continuous rigid body
 I = ∫r2 dm
Parallel axis theorem
R
Rotational inertia about a given axis =
Rotational Inertia about a parallel axis
that extends trough body’s Center of
Mass + Mh2
I  I COM  Mh
2
h = perpendicular distance between the given axis and axis
through COM.
Proof:

( x  a )  ( y  b ) dm 


 ( x  y ) dm  2 a xdm  2b ydm  ( a


 
I  r 2 dm 
2
2
2
2
2
 b 2 ) dm
I   R dm  2 aMx COM  2 bMy COM  Mh  I COM  Mh
2
2
2
Torque
Torque:
Twist  “Turning action of force F ”.
Units: Nm
Radial component, Fr : does not
cause rotation  pulling a door
parallel to door’s plane.
Tangential component, Ft: does cause rotation  pulling a
door perpendicular to its plane. Ft= F sinφ
t  r  ( F  sin  )  r  Ft  ( r sin  ) F  r F
r┴ : Moment arm of F
Vector quantity
r : Moment arm of Ft
Sign: Torque >0 if body rotates counterclockwise.
Torque <0 if clockwise rotation.
Superposition principle: When several torques act on
a body, the net torque is the sum of the individual
torques
Newton’s second law for rotation
F  m a  t  Ia
Proof:
Particle can move only along the
circular path  only the tangential
component of the force Ft (tangent
to the circular path) can accelerate
the particle along the path.
F t  mat
t  Ft  r  mat  r  m (a  r ) r  ( mr 2 )a  Ia
t n et  Ia
VII. Work and Rotational kinetic energy
Translation
DK  K f  K i 
1 2 1 2
mv f  mvi  W
2
2
xf
W   Fdx
Rotation
DK  K f  K i 
1 2 1 2
Iw f  Iwi  W
2
2
qf
W   t  dq
xi
qi
W  F d
W  t (q f  q i )
P
dW
 F v
dt
Work-kinetic energy
Theorem
Work, rotation about fixed axis
Work, constant torque
Power, rotation about
fixed axis
dW
P
 t w
dt
Proof:
W  DK  K f  K i 
1 2 1 2 1
1
1
1
1
1
mv f  mvi  m(w f r ) 2  m(wi r ) 2  (mr 2 )w 2f  (mr 2 )wi2  Iw 2f  Iwi2
2
2
2
2
2
2
2
2
qf
dW  Ft ds  Ft  r  dq  t  dq  W   t  dq
qi
P
dW t  dq

 t w
dt
dt
Rotational Kinetic Energy
We must rewrite our statements of conservation of
mechanical energy to include KEr
Must now allow that (in general):
½ mv2+mgh+ ½ Iw2 = constant
Could also add in e.g. spring PE
Example - Rotational KE
• What is the linear speed of a ball with radius 1 cm
when it reaches the end of a 2.0 m high 30o
incline?
2m
mgh+ ½ mv2+ ½ Iw2 = constant
• Is there enough information?
Example - Rotational KE
mgh+ ½ mv2+ ½ Iω2 = constant
2
2
I Sphere  MR
5
1
1 2
2
So we
Mghi  Mv f   MR 2w 2f
2
2 5
have that:
1 2 1 2 2
ghi  v f  R w f
2
5
The velocity of the centre of mass and the
tangential speed of the sphere are the same,
so we can say that:
vt v
w 
R R
1 2 1 2
2
ghi  v f  v f  0.7v f
2
5
Rearranging for vf:
ghi
9 .8  2
vf 

 5.3m / s
0 .7
0 .7
Example: Conservation of KEr
A boy of mass 30 kg is flung off the edge of a roundabout
(m = 400 kg, r = 1 m) that is travelling at 2 rpm. What is
the speed of the roundabout after he falls off?
Example: Conservation of KEr
A boy of mass 30 kg is flung off the edge of a roundabout
(m = 400 kg, r = 1 m) that is travelling at 2 rpm. What is
the speed of the roundabout after he falls off?
Roundabout is a disk:
Boy has
1
I R  MR 2  0 .5  ( 400 kg )  (1m ) 2  200 kgm 2
2
I B  M B R  ( 30 k g )(1m )  30 k gm
2
2
I total  I R  I B  200 k gm 2  30 k gm 2  230 k gm 2
1
2
2  2 
KEi  I iwi  0.5( 230kgm ) 2
  5J
2
 60 
1
1
KE f  I f w 2f  ( 200kgm2 )w 2f
KEi  KE f
2
2
5
wf 
 0.22rad / s
0.5  200
2
2
During a certain period of time, the angular position of a
swinging door is described by
θ= 5.00 + 10.0t + 2.00t2
where θ is in radians and t is in seconds. Determine the
angular position, angular speed, and angular
acceleration of the door (a) at t = 0 and (b) at t = 3.00 s.
During a certain period of time, the angular position of a
swinging door is described by
θ= 5.00 + 10.0t + 2.00t2
where θ is in radians and t is in seconds. Determine the
angular position, angular speed, and angular
acceleration of the door (a) at t = 0 and (b) at t = 3.00 s.
Solution:
(a)
q
t 0
 5.00 rad
w t 0
dq

 10.0  4.00tt 0  10.0 rad s
dt t 0
a t 0
dw

 4.00 rad s2
dt t 0
During a certain period of time, the angular position of a
swinging door is described by
θ= 5.00 + 10.0t + 2.00t2
where θ is in radians and t is in seconds. Determine the
angular position, angular speed, and angular acceleration of
the door (a) at t = 0 and (b) at t = 3.00 s.
Solution:
(b)
q t 3.00 s  5.00  30.0  18.0  53.0 rad
dq
w t 3.00 s 
 10.0  4.00tt 3.00 s  22.0 rad s
dt t 3.00 s
dw
a t 3.00 s 
 4.00 rad s2
dt t 3.00 s
The four particles are connected by
rigid rods of negligible mass. The
origin is at the center of the
rectangle. If the system rotates in
the xy plane about the z axis with
an angular speed of 6.00 rad/s,
calculate (a) the moment of inertia
of the system about the z axis and
(b) the rotational kinetic energy of
the system.
The four particles are connected by
rigid rods of negligible mass. The
origin is at the center of the
rectangle. If the system rotates in
the xy plane about the z axis with
an angular speed of 6.00 rad/s,
calculate (a) the moment of inertia
of the system about the z2axis and
I  kinetic
m jrenergy
(b) the rotational
of
j
the system.
j

(a)
r1  r2  r3  r4
r
( 3.00 m 
2
 ( 2.00 m
2

2
 13.0 m
I   13.0 m   3.00  2.00  2.00  4.00 kg
 143 kg  m
2
The four particles are connected by
rigid rods of negligible mass. The
origin is at the center of the rectangle.
If the system rotates in the xy plane
about the z axis with an angular speed
of 6.00 rad/s, calculate (a) the moment
of inertia of the system about the z axis
and (b) the rotational kinetic energy of
the system.
(b)
(
1 2 1
K R  Iw  143 kg  m
2
2
 2.57  103 J
In this case,
2
 ( 6.00 rad s
2
Many machines employ cams for various purposes, such as
opening and closing valves. In Figure, the cam is a circular
disk rotating on a shaft that does not pass through the center
of the disk. In the manufacture of the cam, a uniform solid
cylinder of radius R is first machined. Then an off-center hole
of radius R/2 is drilled, parallel to the axis of the cylinder, and
centered at a point a distance R/2 from the center of the
cylinder. The cam, of mass M, is then slipped onto the circular
shaft and welded into place. What is the kinetic energy of the
cam when it is rotating with angular speed ω about the axis of
the shaft?
We consider the cam as the superposition of the original solid disk and a disk of
negative mass cut from it. With half the radius, the cut-away part has one-quarter the
face area and one-quarter the volume and one-quarter the mass of the original solid
cylinder:
1
M 0 M
4
0
M
M
0
4
 M
3
By the parallel-axis theorem, the original cylinder had moment of inertia
2
IC M
1
R
 M 0    M 0R 2  M
 2
2
0
R2 3
 M 0R 2
4
4
The negative mass portion has :
2
The whole cam has:
M 0R 2
1 1
R
I   M 0    
  2
2 4
32
2
M
R
3
23
23 4
23
2
2
2
0
I  M 0R 

M 0R 
MR 
M R2
4
32
32
32 3
24
and
1 2 1 23
23
2 2
K  Iw 
MR w 
M R 2w 2
2
2 24
48
Find the net torque on the wheel in Figure about the axle
through O if a = 10.0 cm and b = 25.0 cm.
Find the net torque on the wheel in Figure about the axle
through O if a = 10.0 cm and b = 25.0 cm.
 t  0.100 m ( 12.0 N   0.250 m ( 9.00 N   0.250 m ( 10.0 N  
The thirty-degree angle is unnecessary information.
 3.55 N  m
A block of mass m1 = 2.00 kg
and a block of mass m2 =
6.00 kg are connected by a
massless string over a pulley
in the shape of a solid disk
having radius R = 0.250m
and mass M = 10.0 kg.
These blocks are allowed to move on a fixed block-wedge
of angle = 30.0 as in Figure. The coefficient of kinetic
friction is 0.360 for both blocks. Draw free-body diagrams
of both blocks and of the pulley. Determine (a) the
acceleration of the two blocks, and (b) the tensions in the
string on both sides of the pulley.
For m1:
 Fy  m ay
 n  m 1g  0
n1  m 1 g  19.6 N
fk1  m k n1  7.06 N
 Fx  m ax
 7.06 N  T1  ( 2.00 kg  a
For pulley,
 t  Ia
1
 T1  T2  ( 10.0 kg  a
2
1
2  a
 T1R  T2R  M R  
 R
2
 T1  T2  ( 5.00 kg  a
For m2:
 n 2  m 2 g cosq  0
(

fk2  m k n2
 18.3 N
n2  6.00 kg 9.80 m s2 ( cos30.0 
 50.9 N
 18.3 N  T2  m 2 sin q  m 2 a
 18.3 N  T2  29.4 N  ( 6.00 kg  a
(a) Add equations for m1,
m2, and for the pulley:
 7.06 N  T1  ( 2.00 kg  a
 T1  T2  ( 5.00 kg  a
 18.3 N  T2  29.4 N  ( 6.00 kg  a
7.06 N  18.3 N  29.4 N  ( 13.0 kg a
4.01 N
a
 0.309 m s2
13.0 kg
(b)
(

T1  2.00 kg 0.309 m s2  7.06 N  7.67 N
(
2

T2  7.67 N  5.00 kg 0.309 m s  9.22 N
A uniform rod 1.1 m long with mass 0.7 kg is pivoted at
one end, as shown in Fig., and released from a horizontal
position. Find the torque about the pivot exerted by the
force of gravity as a function of the angle that the rod
makes with the horizontal direction.
A uniform rod 1.1 m long with mass 0.7 kg is pivoted at
one end, as shown in Fig., and released from a horizontal
position. Find the torque about the pivot exerted by the
force of gravity as a function of the angle that the rod
makes with the horizontal direction.
From the diagram we see
that the moment arm of the
force of gravity is ( 1 L ) cos
2
θ
q.
mg
Thus
t  ( 21 L) cos q mg
 21 (11
. m)(cos q )(0.7 kg)(9.8 m/s2 )  4 cos q .
A seesaw pivots as shown in Fig. (a) What is the net torque
about the pivot point? (b) Give an example for which the
application of three different forces and their points of
application will balance the seesaw. Two of the forces must
point down and the other one up.
A seesaw pivots as shown in Fig. (a) What is the net torque
about the pivot point? (b) Give an example for which the
application of three different forces and their points of
application will balance the seesaw. Two of the forces must
point down and the other one up.
7N
3N
2.5 m
(a)
1.4 m
We choose the clockwise direction as positive.
t p ivo t  ( 7 N )(1 . 4 m )  ( 3 N )( 2 . 5 m )  2 . 3 N  m
(b) We see that changing the 7-N force to ~5.4 N will make the
torque zero. An upward force of 8.4 N at the pivot will make
the resultant force equal to zero.
Four small spheres are fastened to the corners of a frame of
negligible mass lying in the xy plane (Fig. 10.7). Two of the
spheres have mass m = 3.1kg and are a distance a = 1.7 m
from the origin and the other two have mass M = 1.4 kg and
are a distance a = 1.5 m from the origin.
(a) If the rotation of the system occurs about the y axis, as in
Figure a, with an angular speed ω = 5.1rad/s, find the
moment of inertia Iy about the y axis and the rotational kinetic
energy about this axis.
Suppose the system rotates in the xy plane about an axis
(the z axis) through O (Fig. b). Calculate the moment of inertia
about the z axis and the rotational energy about this axis.
Four small spheres are fastened to the corners of a frame
of negligible mass lying in the xy plane (Figure). Two of the
spheres have mass m = 3.1kg and are a distance a = 1.7 m
from the origin and the other two have mass M = 1.4 kg and
are a distance a = 1.5 m from the origin.
(a) If the rotation of the system occurs about the y axis, as
in Figure a, with an angular speed ω = 5.1rad/s, find the
moment of inertia Iy about the y axis and the rotational
kinetic energy about this axis.
Suppose the system rotates in the xy plane about an axis
(the z axis) through O (Fig. b). Calculate the moment of
inertia about the z axis and the rotational energy about this
axis.
(a)
Iy 

mi ri2  2(1.4kg)(1.5m) 2  6.3kg  m 2
i
1
K R  I yw 2  0.5(6.3kg  m 2 )(5.1rad / s ) 2  81.9 J
2
Four small spheres are fastened to the corners of a frame
of negligible mass lying in the xy plane (Fig. 10.7). Two of
the spheres have mass m = 3.1kg and are a distance a =
1.7 m from the origin and the other two have mass M = 1.4
kg and are a distance a = 1.5 m from the origin.
(a) If the rotation of the system occurs about the y axis, as
in Figure a, with an angular speed ω = 5.1rad/s, find the
moment of inertia Iy about the y axis and the rotational
kinetic energy about this axis.
Suppose the system rotates in the xy plane about an axis
(the z axis) through O (Fig. b). Calculate the moment of
inertia about the z axis and the rotational energy about this
axis.
(b)
Iz 

mi ri2  2(1.4kg)(1.5m) 2  2(3.1kg)(1.7 m) 2  24.22kg  m 2
i
1
K R  I zw 2  0.5(24.22kg  m2 )(5.1rad / s ) 2  315.0 J
2
The reel shown in Figure has
radius R and moment of
inertia I. One end of the block
of mass m is connected to a
spring of force constant k, and
the other end is fastened to a
cord wrapped around the reel.
The reel axle and the incline
are frictionless. The reel is
wound counterclockwise so
that the spring stretches a
distance d from its unstretched position and is then released
from rest. (a) Find the angular speed of the reel when the
spring is again unstretched. (b) Evaluate the angular speed
numerically at this point if I = 1.00 kg·m2, R = 0.300 m,
k = 50.0 N/m, m = 0.500 kg, d = 0.200 m, and θ= 37.0°.
(a)
W  DK  DU
W

D
K

D
U
W  K  K U U
f
i
f
W  K f  K i U f  U i
1 2 1 2
1 2
0  m v  Iw  m gd sin q  kd
2
2
2
1 2
1 2
2
w I m R  m gd sin q  kd
2
2
(
w

2m gd sin q  kd
2
I m R
2
i
w
2 mgd sin q  ka
2
I  mR 2
(b)
w
(
( sin 37.0  50.0 N

2
1.00 kg  m 2  0.500 kg ( 0.300 m 
2( 0.500 kg  9.80 m s2 ( 0.200 m
w
m ( 0.200 m
1.18  2.00
 3.04  1.74 rad s
1.05
2
A tennis ball is a hollow sphere with a thin wall. It is set
rolling without slipping at 4.03 m/s on a horizontal section of
a track, as shown in Figure. It rolls around the inside of a
vertical circular loop 90.0 cm in diameter, and finally leaves
the track at a point 20.0 cm below the horizontal section.
(a) Find the speed of the ball at the top of the loop.
Demonstrate that it will not fall from the track. (b) Find its
speed as it leaves the track. (c) Suppose that static friction
between ball and track were negligible, so that the ball slid
instead of rolling. Would its
speed then be higher, lower,
or the same at the top of the
loop? Explain.
A tennis ball is a hollow sphere with a thin wall. It is set rolling without
slipping at 4.03 m/s on a horizontal section of a track, as shown in
Figure. It rolls around the inside of a vertical circular loop 90.0 cm in
diameter, and finally leaves the track at a point 20.0 cm below the
horizontal section. (a) Find the speed of the ball at the top of the loop.
Demonstrate that it will not fall from the track. (b) Find its speed as it
leaves the track. (c) Suppose that static friction between ball and track
were negligible, so that the ball slid instead of rolling. Would its speed
then be higher, lower, or the same at the top of the loop? Explain.
(a) Energy conservation for the system of the ball and the Earth between the
horizontal section and top of loop:
1 2 1 2
1
1
m v2  Iw 2  m gy2  m v12  Iw12
2
2
2
2
1 2
m v2 
2
2
1  2 2   v2 
 m r     m gy2
2 3
r
1
 m v12 
2
5 2
5 2
v2  gy2  v1
6
6
1  2 2   v1 
 m r   
2 3
r
2
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on a
horizontal section of a track, as shown in Figure. It rolls around the inside of a vertical circular
loop 90.0 cm in diameter, and finally leaves the track at a point 20.0 cm below the horizontal
section. (a) Find the speed of the ball at the top of the loop. Demonstrate that it will not fall
from the track. (b) Find its speed as it leaves the track. (c) Suppose that static friction between
ball and track were negligible, so that the ball slid instead of rolling. Would its speed then be
higher, lower, or the same at the top of the loop? Explain.
(a) Energy conservation for the system of the ball and the Earth between the
horizontal section and top of loop:
1 2 1 2
1
1
m v2  Iw 2  m gy2  m v12  Iw12
2
2
2
2
1 2
m v2 
2
2
1  2 2   v2 
 m r     m gy2
2 3
r
1 2
 m v1 
2
5 2
5
v2  gy2  v12
6
6
v2  v12 
6
gy2 
5
1  2 2   v1 
 m r   
2 3
r
2
( 4.03 m s2  5 (9.80 m
6

s2 ( 0.900 m   2.38 m s