physics 231 lecture 20: equilibrium and more rotations
DESCRIPTION
PHYSICS 231 Lecture 20: equilibrium and more rotations. Remco Zegers Walk-in hour: Monday 9:15-10:15 am Helproom. gravitation. Only if an object is near the surface of earth one can use: F gravity =mg with g=9.81 m/s 2 In all other cases: - PowerPoint PPT PresentationTRANSCRIPT
PHY 2311
PHYSICS 231Lecture 20: equilibrium and more
rotations
Remco ZegersWalk-in hour: Monday 9:15-10:15 am
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 r: radius of planetT: period(time to make one rotation) of planet
Our solar system!
PHY 2313
A child of 40 kg is sitting in a Ferris wheel, rotating with an angular velocity of 0.4 rad/s. The radius of the wheelis 9 m. What is the force exerted by the seat on the childat the top and at the bottom.
top: Fcenter=0 -mg+n+mac=0 n=mg-m2r=392-58=334bottom: Fcenter=0 -mg+n-mac=0 n=mg+m2r=392+58=450
The seat is fixed to the wheel. Its centripetal accelerationis directed towards the center.Top: the seat is accelerated away from the childBottom: the seat is accelerated towards the child
PHY 2314
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 2315
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 2316
quiz (extra credit)
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 2317
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 2318
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 2319
Demo: fighting sticks
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