PHY 231 1 PHYSICS 231 Lecture 18: equilibrium and more rotations Remco Zegers Walk-in hour: Tuesday...
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Transcript of PHY 231 1 PHYSICS 231 Lecture 18: equilibrium and more rotations Remco Zegers Walk-in hour: Tuesday...
PHY 2311
PHYSICS 231Lecture 18: equilibrium and more
rotations
Remco ZegersWalk-in hour: Tuesday 4:00-5:00 pm
Helproom
PHY 2312
gravitationOnly if an object is near the surface of earth one can use:Fgravity=mg with g=9.81 m/s2
In all other cases:
Fgravity= GMobjectMplanet/r2 with G=6.67E-11 Nm2/kg2
This will lead to F=mg but g not equal to 9.8 m/s2 (see Previous lecture!)
If an object is orbiting the planet:
Fgravity=mac=mv2/r=m2r with v: linear velocity =angular vel.
So: GMobjectMplanet/r2 = mv2/r=m2r Kepler’s 3rd law: T2=Ksr3 Ks=2.97E-19 s2/m3
T: period(time to make one rotation) of planet r: distance object to planet
Our solar system!
PHY 2313
Previously
Translational equilibrium: F=ma=0 The center of gravitydoes not move!
Rotational equilibrium: =0 The object does notrotate
Mechanical equilibrium: F=ma=0 & =0 No movement!
Torque: =Fd
ii
iii
CG m
xmx
ii
iii
CG m
ymyCenter of
Gravity:
Demo: Leaning tower
PHY 2314
examples: A lot more in the book!
Where is the center of gravity?
0067.018
12.0
1116
)53cos(1.01)53cos(1.01016 00
ii
iii
CG m
xmx
01116
)53sin(1.01)53sin(1.01016 00
ii
iii
CG m
ymy
PHY 2315
question
0-2 1 2 m
20N 40N
40N
A wooden bar is initially balanced. Suddenly, 3 forces areapplied, as shown in the figure. Assuming that the bar canonly rotate, what will happen (what is the sum of torques)?a) the bar will remain in balanceb) the bar will rotate counterclockwisec) the bar will rotate clockwise
Torque: =Fd
PHY 2316
Weight of board: wWhat is the tension in each of thewires (in terms of w)?
w
T1 T2
0
Translational equilibriumF=ma=0T1+T2-w=0 so T1=w-T2
Rotational equilibrium
=0T10-0.5*w+0.75*T2=0T2=0.5/0.75*w=2/3w T1=1/3w
T2=2/3w
PHY 2317
s=0.5 coef of friction between the wall andthe 4.0 meter bar (weight w). What is the minimum x where you can hang a weight w for which the bar does not slide?
w
sn
n
w
T Ty
Tx
(x=0,y=0)
Translational equilibrium (Hor.)Fx=ma=0n-Tx=n-Tcos37o=0 so n=Tcos37o
Translational equilibruim (vert.)Fy=ma=0sn-w-w+Ty=0sn-2w+Tsin37o=0sTcos370-2w+Tsin370=01.00T=2w Rotational equilibrium:
=0xw+2w-4Tsin370=0 so w(x+2-4.8)=0x=2.8 m
PHY 2318
another example
12.5N
30N
0.2L
0.5L
L
Does not move!
What is the tension in the tendon?
Rotational equilibrium:
T=0.2LTsin(155o)=0.085LT
w=0.5L*30sin(40o)=-9.64L
F=L*12.5sin(400)=-8.03L
=-17.7L+0.085LT=0T=208 N
PHY 2319
Demo: fighting sticks
UNTIL HERE FOR MIDTERM II !!!! (until 8.4 in the book)
PHY 23110
rFt=mat
Torque and angular acceleration
m
FNewton 2nd law: F=ma
Ftr=mrat
Ftr=mr2 Used at=r
=mr2 Used =Ftr
The angular acceleration goes linear with the torque.
Mr2=moment of inertia
PHY 23111
Two masses
r
m
F
m r
=mr2=(m1r1
2+m2r22
)
If m1=m2 and r1=r2
=2mr2Compared to the case with only one mass, the angularacceleration will be twice smaller when applying the same torque, if the mass increases by a factor of two.The moment of inertia has increased by a factor of 2.
PHY 23112
Two masses at different radii
r
m
F
mr
=mr2=(m1r1
2+m2r22
)
If m1=m2 and r2=2r1
=5mr2When increasing the distance between a mass and therotation axis, the moment of inertia increases quadraticly.So, for the same torque, you will get a much smallerangular acceleration.
PHY 23113
A homogeneous stick
Rotation point
mmmm
mmmm
m
m
F =mr2=(m1r1
2+m2r22+…+mnrn
2)=(miri
2)=I
Moment of inertia I:
I=(miri2)
PHY 23114
Two inhomogeneous sticks
mmmm
mmmm
5m
5m
F5mmmm
mmm5m
m
m
F
18m 18 m
=(miri2)
118mr2=(miri
2) 310mr2
r
Easy to rotate! Difficult to rotate
PHY 23115
More general.
=IMoment of inertia I:I=(miri
2)
compare with:F=maThe moment of inertiain rotations is similar tothe mass in Newton’s 2nd law.
PHY 23116
A simple example
A and B have the same total mass. If the sametorque is applied, which one accelerates faster?
FF
r r
Answer: A=IMoment of inertia I:I=(miri
2)
PHY 23117
The rotation axis matters!
I=(miri2)
=0.2*0.52+0.3*0.52+ 0.2*0.52+0.3*0.52
=0.5 kgm2
I=(miri2)
=0.2*0.+0.3*0.52+ 0.2*0+0.3*0.52
=0.3 kgm2