Equations Angular displacement ɵ ɵ final ɵ initial ɵ arc length s r 2 п 1 full revolution 2 п r circumference Angular velocity ᾠ ɵ t Angular Acceleration ID: 733420
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Slide1
Physics 111Rotational Motion + inertiaSlide2
Equations:
Angular displacement
ɵ = ɵ final - ɵ initial
ɵ = arc length (s) / r
2
п
= 1 full revolution
2
п
r = circumference
Angular velocity
ᾠ
= ɵ / t
Angular Acceleration
ᾰ
=
ᾠ
/ t
Tangential Velocity (
Vt
) = r (
ᾠ
)
Tangential Acceleration (At) = r(
ᾰ
)
Centripetal Acceleration (Ac) = r (
ᾠ
^2)
Centripetal Force (
Fc
) =
mr
ᾠ
^2
Impluse
(I) = mr^2
Kinetic Energy = .5 I
ᾠ
^2
ᾠ
f =
ᾠ
i
+
ᾰ
t
ɵ = .5 (
ᾠ
i
+
ᾠ
f) t
Ɵf
=
ɵi
+
ᾠ
i
(t) + .5
ᾰ
t^2
ᾠ
f^2 =
ᾠ
i^2+ 2
ᾰ
ɵSlide3
Practice problem:
A motor slows from 34
rad
/s to 10
rad
/s while the turning the drum through an angle of 323 radians. What is the magnitude of the angular acceleration of the motor?Slide4
Solution:
ωF
^
2 = ω0
^
2 + 2αΔθ
10^2 = 34^2 + 2(
α
)(323)
Α
= -1.63
rad
/sec62Slide5
More problems
Passengers are seated in a horizontal circle of radius 4 m. The seats begin from rest and are uniformly accelerated for 13 seconds to a maximum rotational speed of .6
rad
/s.
What is the tangential acceleration of the passengers during the first 30 s of the ride?Slide6
Solution:
At = r
ᾰ
ᾠ
f =
ᾠ
i
+
ᾰ
t
ᾰ
=
ᾠ
f /t =.6 / 13 = .046
rad
/ sec ^2
At = r
ᾰ
= 4* .046 = .185 m/sec^2
Vt
= r
ᾠ
ᾠ
f =
ᾠ
i
+
ᾰ
t
ᾠ
= .046 * 30 sec = 1.38
rad
/s
Vt
= r
ᾠ
= 4 * 1.38 = 5.52 m/s Slide7
Problem Part1:
A bicycle wheel of radius
r
= 1.5
m
starts from rest and rolls 100 m without slipping in 30 s. Calculate
a) the number of revolutions the wheel makes
b) the number of radians through which it turns
c)The average angular velocity.Slide8
Solution:
a)
If there is no slipping, the arc length through which a point of the rim moves is equal to the distance travelled, so that the number of revolutions is:
n
= 100 m / 2
п
(1.5)
= 10.6
b)
Delta
ɵ
= 2
п
n
= s/r =100m/1.5 = 66.7
radians
.
c)
Average angular velocity: = Delta
ɵ / t
= 66.7 /
30
* .5 seconds
=
4.44
rads
/
s
.Slide9
Part 2:
Assuming that the angular acceleration of the wheel given above was constant, calculate:
a) The angular acceleration
b) the final angular velocity
c) the tangential velocity and tangential acceleration of a point on the rim after one revolution.Slide10
Solution:
a)
For constant angular acceleration:
Ɵf
=
ɵi
+
ᾠ
i
(t) + .5
ᾰ
t^2
Using
ᾠ
i
= 0 and solving for
ᾰ
gives: = 2(66.7)/(30^2) = .15
rad
/ s^2
b)
ᾠ
f =
ᾠ
i
+
ᾰ
t
0 + (0.15)(30) = 4.5
rads
/
s
c)
After one revolution, delta
Ɵ
= 2
п
ᾠ
f^2 =
ᾠ
i^2+ 2
ᾰ
ɵ
1.37
rads
/
s
.
The tangential velocity and acceleration are:
Vt
= r
ᾠ
=
(1.5
m
)(1.37
rads
/
s
) = 2.06
m
/
s
a
t
=
r
ᾰ
= (1.5
m
)(0.15
rads
/
s
2
) = .225
m
/
s
2
.