Chapter 10: Dynamics of Rotational Motion • What causes an angular acceleration?
The effectiveness of a force at causing a rotation is called torque.
The four forces shown have the same strength. Which force would be most effec9ve in opening the door?
A. Force F1. B. Force F2. C. Force F3. D. Force F4. E. Either F1 or F3.
QuickCheck 12.5
Slide 12-‐59
Slide 12-‐63
Torque The ability of a force to cause a rotation depends on
1. the magnitude F of the force.
2. the distance r from the point of application to the pivot.
3. the angle at which the force is applied.
Slide 12-‐65
Torque • The magnitude of the torque is given by
Torque is often calculated using the moment arm or lever arm.
|~⌧ | = rF sin(�)
|~⌧ | = r?F = dF
Torque and Cross Product (Section 1.10) Technically, torque is a vector, defined as
| ~A⇥ ~B| = AB sin�
~⌧ = ~r ⇥ ~F
A plumber pushes straight down on the end of a long wrench as shown. What is the magnitude of the torque he applies about the pipe at lower right?
A. (0.80 m)(900 N) sin 19°
B. (0.80 m)(900 N) cos 19°
C. (0.80 m)(900 N) tan 19°
D. none of the above
Q10.3
Clicker Question
QuickCheck 12.6
Slide 12-‐66
Clicker Question
• Which third force on the wheel, applied at point P, will make the net torque zero?
Torque and Angular Acceleration
• Consider rocket (of mass m) attached to a massless rod (with no other forces acting on it).
• The net tangential force causes a tangential acceleration:
• Multiplying both sides by r: • Expressing as vectors:
Ft = mat = mr↵
⌧ = rFt = mr2↵ = I↵
~⌧ = I ~↵• For a constrained axis of rota9on, we only use component of
torque that is along the axis.
Torque and Angular Acceleration
• Now let’s generalize by considering a rigid body made up of many particles.
• Net tangential force on particle 1:
• Summing over all particles:
– Torques due to internal forces cancel out due to Newton’s 3rd law.
– Net torque only due to external forces.
Ft1 = m1at1⌧z1 = r1Ft1 = m1r
21↵
X
i
⌧zi =X
i
riFti =X
mir2i ↵
⌧netz = I↵z
Clicker Question
Chalkboard Question A 40 kg door has a height of 2 m, a width of 1.5 m, and
a door knob 0.1 m from the edge. A person exerts a force of 50 N on the door knob, directed 30 degrees with respect to the normal of the door. What is the resulting angular acceleration and estimate how long it takes for the door to open 45 degrees.
Clicker Question
A. m2g = T2 = T1
B. m2g > T2 = T1
C. m2g > T2 > T1
D. m2g = T2 > T1
E. none of the above
Q10.5
A glider of mass m1 on a frictionless horizontal track is connected
to an object of mass m2 by a massless string. The glider
accelerates to the right, the object accelerates downward, and the
string rotates the pulley. What is the relationship among T1 (the
tension in the horizontal part of the string), T2 (the tension in the
vertical part of the string), and the weight m2g of the object?
2
T1
T
Net torque = (T1 – T2) R 6
T1
T
T TTT
2 34
5
⌧ = (T1 � T2 + T2 � T3 + T3 � T4 + T4 � T5 + T5 � T6)R⌧ = (T1 � T6)R
Chalkboard Question • What is acceleration of 125 N weight? The pulley has
a mass of 5 kg and a radius of 0.3 m (solid disk).
Torque Due to Gravity
• Gravity acts on each chunk of mass. Net torque given by
• Can treat net torque as due to gravity acting on center of mass
⇥� =�
i
⇥ri ⇥ (�migj)
~⌧ = ~RCM ⇥ (�MT g j)
4
x
y
r4
m4
F = m g
Mg
x
y
RCM
~⌧ =
Pi(mi~ri)
MT
�⇥ (�MT g j)
Angular Acceleration of a Falling Tree • Determine the angular acceleration of a falling tree
(just as the roots let go). Treat the tree as a uniform rod of length L and mass M.
From HewiW’s Conceptual Physics
• Example: Teeter Toter
• If M2r2 = M1r1, net torque is zero. Notice that this means the pivot point is the center of mass!
�net = M2gr2 �M1gr1
Combined Translation and Rotation: Dynamics • For an object that doesn’t have a fixed axis of rotation, the
object can be analyzed by looking at the rotational dynamics with respect to the center of mass AND looking at the motion of the center of mass (due to the net force).
~⌧ = Icm ~↵
~Fnet = MT ~acm
Clicker Question
A yo-‐yo is placed on a horizontal surface as shown. There is sufficient fric9on for the yo-‐yo to roll without slipping. If the string is pulled to the right as shown,
A. the yo-‐yo rolls to the right.
B. the yo-‐yo rolls to the le].
C. the yo-‐yo remains at rest. D. The answer depends on the magnitude F of the pulling force compared to the magnitude of the fric9on force.
Q10.9
F
Clicker Question
Clicker Question
A solid bowling ball rolls down a ramp.
Which of the following forces exerts a torque on the bowling ball about its center?
A. the weight of the ball
B. the normal force exerted by the ramp
C. the friction force exerted by the ramp
D. more than one of the above
E. The answer depends on whether the ball rolls without slipping.
Q10.7
Rolling Down an Incline • Determine the acceleration of uniform spheres, rings,
and cylinders as they roll down an incline of slope θ (assuming they don’t slip).
Which will win?
Chalkboard Question • Impulse J given to ball by exerting a force F with cue
at a height h above center of mass. For what value of h will the ball roll without slipping (assuming F>>force of static friction)? If h=0, how far does it slide before rolling without slipping, and how fast is it moving once it stops skidding?
Chalkboard Question A bowler throws a bowling ball of radius R=11 cm
down a lane. The ball slides on the lane, with initial speed 8 m/s and initial angular velocity ω0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.33. How long and how far does it take for the ball to begin rolling smoothly without skidding?
Chalkboard Question A person of mass mp=75 kg is on a unicycle of mass mu
= 1 kg. The radius of the unicycle wheel is Rw = 1.2 m and the pedal arm has a length Rp = 0.3 m. The person can push down on the pedal with a force F = 750 N. Assuming the wheel doesn’t slip on the ground, what is the acceleration of the unicycle?
Work and Power in Rotational Motion • Consider again pure rotational motion, where a force
is causing an angular displacement. The work can be determined using
• The power is given by
W =�
⇤F · ⇤ds =�
FtanR d� =� �2
�1
⇥ d�
P =dW
dt= � ⇥
Example • A certain car can produce 200 hp at 6,000 rpm. What
is the torque the engine is exerting at this point?
P = 200 hp� 746 W/hp = 1.5� 105 W
⇥ = 2� � 6, 000 rpm = 628 rad/s
� = P/⇥ = 240 N m
Angular Momentum • Consider a single point-mass object of mass m,
moving a distance r from some arbitrary origin. Define angular momentum as
• What is
• Thus:
• If the torque is zero, then angular momentum is conserved.
�L = �r � �p
d�L
dt= �v � �p + �r � d�p
dt= �r � �F
d⇥L
dt= ⇥�
d~L
dt?
Angular Momentum of a Rigid Body • For a rotating rigid body, the total angular momentum
is simply the sum of the individual angular momentum of each mass element:
• If the torque is zero, and the moment of inertia changes, then the angular velocity must change such that L remains constant in time.
�L =�
i
�Li
L =⇤
i
Li = (ri)(miri�) =
�⇤
i
mir2i
⇥� = I�
⇥z =dLz
dt= I�z
I! = I 0!0
A star is rotating with a period T. Over a period of a million years, its radius decreases by a factor of 2 (without losing any mass) . What is the new period of the star? (Hint: ) I M Rsphere =
25
2
A)T/2 B) 2T C) 4T D) T/4 E) None of these.
Clicker Question
(from CU Boulder’s PER site)
A spinning figure
skater pulls his arms
in as he rotates on
the ice. As he pulls
his arms in, what
happens to his
angular momentum L
and kinetic energy K?A. L and K both increase
B. L stays the same, K increases
C. L increases, K stays the same
D. L and K both stay the same
Q10.11
JITT Question • Suppose you are standing on the edge of a spinning
merry-go-round. You step off, at right angles to the edge. What effect (if any) does this have on the angular speed of the merry-go-round?
HW Question: • How fast is CM right before it lands?
Example 10.14: A 10 g bullet is shot into the center of a 15 kg, 1 m wide
door. The bullet is shot with a speed of 400 m/s. Find the angular speed of the door after the bullet embeds itself in the door.
Gyroscopes and Precession
Static Equilibrium (11.1-11.3) • The conditions for static equilibrium are:
�Fnet = 0
⇥�net = 0
Torque Due to Gravity
• Gravity acts on each chunk of mass. Net torque given by
• Can treat net torque as
due to gravity acting on center of mass
⇥� =�
i
⇥ri ⇥ (�migj)
⇥� =�
i
(mi⇥ri)⇥ (�gj) = ⇥RCM ⇥ (�Mg j)
JITT Comments • Sometimes the reading said center of gravity, other
times center of mass. I thought they were the same...? • The book insinuates that for the most part center of
gravity and center of mass are the same thing. When are they different? For the purposes of our class, do we treat them the same?
• Up until now, we have not been considering the torque of gravity. Does the center of gravity equal the center of mass only when a mass is symmetrical?
• Does force of gravity always act of the center of mass even if it is not the center of gravity?
• Example: Teeter Toter
• If M2r2 = M1r1, net torque is zero. Notice that this means the pivot point is the center of mass.
�net = M2gr2 �M1gr1
• What force F2 is required to apply a 400 N force to the nail?
Tipping a Box Over
• How hard do you have to push on a box so it tips over (assume it doesn’t slip)
F
h
a
b
Chalkboard Question A ladder of length L and mass M is placed against a
frictionless wall. The coefficient of static friction between the ladder and the ground is 0.7. What is the minimum angle between the ladder and the ground such that the ladder doesn’t slip and fall?
Tipping a Box Over • How tall does the box have to be for it to tip over
while sliding to a stop?
h
a
b v
Why Lean when Turning on a Bike? • Consider a wheel of radius R rolling along the
ground, making a circle of radius r.
• Consider rotating frame of reference such that CM of wheel is stationary.
Torque about contact point:
tan � =v2
gr⇥net = mgR sin � � mv2
rR cos � = 0