angle of rotation ( rads )
DESCRIPTION
Clicker Question. Room Frequency BA. angle of rotation ( rads ). angular velocity ( rad/s ). angular acceleration (rad/s 2 ). Recall that centripetal acceleration is expressed in terms of tangential velocity as: a r = v 2 /r. How is it expressed in terms of angular velocity ω ? - PowerPoint PPT PresentationTRANSCRIPT
atan
r
vr
2
2 f
angle of rotation (rads)
angular velocity (rad/s)
angular acceleration (rad/s2)
Recall that centripetal acceleration is expressed in terms of tangential velocity as: ar = v2/r. How is it expressed in terms of angular velocity ω?
A) ar = ω2/r B) ar = rω
C) ar = rω2 D) ar = r2ω2
ar = v2/r = (rω)2/r = rω2
Clicker Question Room Frequency BA
Remember: This Week
• Read Ch. 8: And read Dr. Dubson’s excellent notes on Ch. 8.
• CAPA assignment #10 is due this Friday at 10 PM.
• Recitation: Review 1
• Lab Make-up (Labs 1-3): arrange with your TA.
atan
r
vr
2
2 f
IC
angle of rotation (rads)
angular velocity (rad/s)
angular acceleration (rad/s2)
torque (N m)moment of inertia (kg m2)Newton’s 2nd Law
Today:
KE PE constant KEtrans KErot PE constant
(Conservation of Mechanical Energy)
Rotational Kinematics
atan
r
vr
2
2 f
constant
angle of rotation (rads)
angular velocity (rad/s)
angular acceleration (rad/s2)
v v0 at
x x0 vot 12at 2
v2 v02 2ax
0 t
0 ot 12t 2
2 02 2
Constant acceleration a: Constant angular acceleration α:
Rotational Kinematics
0 t
vr
2
2 f
0 ot 12t 2
2 02 2
A space ship is initially rotating on its axis with an angular velocity of ω0 = 0.5 rad/sec. The Captain creates a maneuver that produces a constant angular acceleration of α = 0.1 rad/sec2 for 10 sec.
1) What is its angular velocity at the end of this maneuver?
A) 1 rad/sec B) 1.5 rad/sec
C) 0.6 rad/sec D) 5 rad/sec
0 t 0.5 rad/sec (0.1 rad / sec2 )(10sec) 1.5 rad/sec
Clicker Question Room Frequency BA
A space ship is initially rotating on its axis with an angular velocity of ω0 = 0.5 rad/sec. The Captain creates a maneuver that produces a constant angular acceleration of α = 0.1 rad/sec2 for 10 sec.
1) What is its angular velocity at the end of this maneuver? 1.5 rad/sec
f 2revols
sec
1.52revols
sec60secmin
45revolsmin
45rpm
Clicker Question Room Frequency BA
0 t
vr
2
2 f
0 ot 12t 2
2 02 2
2) At how many rpms is it rotating after the maneuver?
A) 10 rpm B) (30/π) rpm C) (45/π) rpmD) π rpm
3) How far (in radians) did the space ship rotate while the Captain was applying the angular acceleration?
A) 1 rad B) 5 rad C) 10 rad D) 45 rad
1.5 rad/sec
f 45rpm
0 ot 12t 2 = (0.5 rad/sec)(10 sec) +
12
(0.1 rad/sec2 )(10 sec)2
= 5 rad +5 rad = 10 rad
Clicker Question Room Frequency BA
0 t
vr
2
2 f
0 ot 12t 2
2 02 2
A space ship is initially rotating on its axis with an angular velocity of ω0 = 0.5 rad/sec. The Captain creates a maneuver that produces a constant angular acceleration of α = 0.1 rad/sec2 for 10 sec.
1) What is its angular velocity at the end of this maneuver?
2) At how many rpms is it rotating after the maneuver?
4) How many degrees is this?
A) 180° B) 1800°
C) (1800/π) degrees D) 360°
1.5 rad/secf
45rpm
= 10 rad
10rad180 deg rad
1800
deg
Clicker Question Room Frequency BA
0 t
vr
2
2 f
0 ot 12t 2
2 02 2
A space ship is initially rotating on its axis with an angular velocity of ω0 = 0.5 rad/sec. The Captain creates a maneuver that produces a constant angular acceleration of α = 0.1 rad/sec2 for 10 sec.
1) What is its angular velocity at the end of this maneuver?
2) At how many rpms is it rotating after the maneuver?
3) How far (in radians) did the space ship rotate while the Captain was applying the angular acceleration?
atan
r
vr
2
2 f
IC
angle of rotation (rads)
angular velocity (rad/s)
angular acceleration (rad/s2)
torque (N m)moment of inertia (kg m2)Newton’s 2nd Law
Today:
KE PE constant KEtrans KErot PE constant
Rotational Dynamics
r
rθ
s F
F
A
Consider the work done by a constant force in moving an object from point A to point B on a circle.
F
W
= Fs F(r) = (rF )
(Component of force along displacement) x displacement
rF Units: N mTorque
Torque Rotational Dynamics
Torques are related to F
Torque:
rF
= (lever arm) x (Component of force perpendicular to lever arm)
The amount of work done by increases linearly with r and .F
F
Torque
F
rF
Therefore, to easily rotate an object about an axis, you want a large lever arm r and a large perpendicular force :
Torque
F
Therefore, to easily rotate an object about an axis, you want a large lever arm r and a large perpendicular force :
rFTorque
What’s the torque you exerted on the door? (Note: sin 30° =1/2)
A) 8 Nm B) 10 Nm
C) 12 Nm D) 20 Nm
rF rF sin (1m)(20N )(1 / 2) 10Nm
rF
Clicker Question Room Frequency BA
Torque
You pull on a door handle a distance r = 1m from the hinge with a force of magnitude 20 N at an angle θ =30° from the plane of the door.
Three forces labeled A, B, and C are applied to a rod which pivots on an axis through its center.
Which force causes the largest size torque?sin 45 cos 45 1 / 2 0.707
| A |LF sin 45 0.707LF
| B |L2F 0.5LF
| C |L4
2F 0.5LF
Clicker Question Room Frequency BA
Torque rF
Three forces labeled A, B, and C are applied to a rod which pivots on an axis through its center.
What is the net torque on the rod?
A) 0.707 LF B) -0.707 LF C) 1.707 LF D) -1.707 LF
E) Zero
A LF sin 45 0.707LF
B L2F 0.5LF
C L4
2F 0.5LF
net (0.707 0.5 0.5)LF 1.707LF
Clicker Question Room Frequency BA
Torque rF
Torque and Moment of Inertia
Rotational Analog to Newton’s Second Law (F = ma):
I
Torque = (moment of inertia) x (rotational acceleration)
Moment of inertia I depends on the distribution of mass and, therefore, on the shape of an object.
Units of τ: NmUnits of I: kg m2
Units of α: rad/s2
rF
Ipoint mr2Moment of inertia for a point mass is the mass times the square of the distance of the mass from the axis of rotation.
Torque and Moment of Inertia
Consider a point mass m to which we apply a constant tangential force .m
r
r F
atan rF matan mr
rF mr2 I
I mr2
Thus, the moment of inertia for apoint mass is equal to mr2.
rF I
A light rod of length 2L has two heavy masses (each with mass m) attached at the end and middle. The axis of rotation is at one end. What is the moment of inertia about the axis?
A) mL2 B) 2mL2 C) 4mL2 D) 5mL2 E) 9mL2
I miri2 = mL2 m(2L)2 5mL2
up
down
rF I
Ipoint mr2
Clicker Question Room Frequency BA
Rotational Dynamics
atan
r
vr
2
2 f
IC
angle of rotation (rads)
angular velocity (rad/s)
angular acceleration (rad/s2)
torque (N m)
moment of inertia (kg m2)
Newton’s 2nd Law
Tod
KE PE constant KEtrans KErot PE constant(Conservation of Mechanical Energy)
ptot mii vi constant Ltot Ii
i i constant
Kinetic Energy (joules J)