Transcript Physics 130 - University of North Dakota

11/6 Torque in Equilibrium
Today: Examples with Torque
HW: 11/6 HW Handout “Torque Disk” due
Friday 11/8 (None due Wed)
Exam 3: Thursday, 11/14
Monday holiday-no lab next week!
Rotational Kinetic Energy intro
Example:
Now that we know the tension in the angled string, we can find
the pivot force components.
NP,B = 61N (rounded) @ 4.7°
NP,B
NP,y = 5N
1kg
T1 = 70N
30°
NP,x = 61N
2m
3kg

CW
=

CCW
W = 10N
=0
10 + 30 = 35 + NP,y
FUp = FDown a = 0
70 cos30° = NP,x
FLeft = FRight a = 0
T1, =
35N
T2 = 30N
Sample problem:
1 kg meter stick
2 kg
m=?
The mass of the meter stick is 1 kg.
Find the mass of the other block.
Rotational Kinetic Energy
Just another bucket
KER
We will use proportional reasoning to
decide how to split up the total KE into
KER and KET energy.
KET is our new name for KE, the T stands
for “translation.”
Rotational Kinetic Energy
Just another bucket
KER
KE = 1/2mv2
Ring has same m and same v
so KER,Ring = KET,Block
Disk has same m but mass near
center is slower so less KER here
KER,Disk = 1/2KET,Block
Ring of mass m
rotating with
velocity v
Block of mass m
moving with
velocity v
Disk of mass m
rotating with
velocity v
0
R
0
T
0
g
S
h
R
T
vT
Rolls without slipping so vR =
vT and KER = KET since it is a
ring.
Use Energy to fins the velocity
of the ring.
0
g
0
S
1/ PE
2
g
= KET
1/ mgh
2
= 1/2mv2
vT = gh
For the disk
0
R
0
T
g
0
S
h
R
Rolls without slipping so vR =
vT and KER = 1/2KET since it is
a disk.
T
0
g
0
S
v=
2/ PE =
3
g
2/ mgh
3
4gh
3
KET
= 1/2mv2