Torque again ...
Torque again ... Sir Lancelot on a ladder (or Example 9-4 in text): Ladder weight = 180 N Sir L’s weight = 800 N
Ladder length = 5 m Sir L is 1/3 of the way up
No friction at wall !
Torque again ... Sir Lancelot on a ladder (or Example 9-4 in text): Free-body diagram of 5.0-m long ladder:
Rotation axis
Torque again ... Sir Lancelot on a ladder (or Example 9-4 in text): Ladder’s weight: τa = −(180 N)(1.5 m) = −270 N · m Sir Lancelot’s normal force: τb = −(800 N)(1.0 m) = −800 N · m Ground’s normal force: τ2 = +(n2 )(0.0 m) = 0 N · m Ground’s friction: τf = +(fs )(0.0 m) = 0 N · m Wall’s normal force: τ1 = +(n1 )(4.0 m) = 4n1
Torque again ... Sir Lancelot on a ladder (or Example 9-4 in text): Rotational equilibrium ! ⇒ τ =0 ⇒ τa + τb + τ2 + τf + τ1 = 0 ⇒ 4n1 = (270 + 800) N · m ⇒ n1 = 267.5 N
Translational equilibrium !
⇒ Fx = 0 ⇒ fs = n1 = 267.5 N ! and Fy = 0 ⇒ n2 = wa + wb = 980 N
Rotational Equilibrium ~ Building a mobile:
Find this Use
!
τ =0
– but first, choose a rotation axis!
Rotational Equilibrium ~ Building a mobile:
Rotation axis τm = +(mg)(0.10 m) τa = −(0.10g N)(0.15 m) τb = −(0.10g N)(0.40 m) τT = +(T )(0.0 m)
!
τ =0 ⇒ τm + τa + τb + τT = 0 ⇒ (mg)(0.1) = (0.1g)(0.55) ⇒ m = 0.55 kg
Rotational Equilibrium w , warm , θ , and all distances are given. ( θ is the angle between T and the horizontal)
Find T , Ex and Ey .
Rotational Equilibrium Rotation axis
Rotational Equilibrium Dumbbell: τd = +(w)(L) Forearm’s weight: τa = +(warm )(L/2) Horizontal component of T: τx = +(Tx )(0) = 0 Vertical component of T: τy = −(Ty )(D) Horizontal component of E: τh = +(Ex )(0) = 0 Vertical component of E: τv = +(Ey )(0) = 0
Rotational Equilibrium !
τ =0 L warm 2
⇒ wL + − Ty D# = $ 0 ! " L 1 ⇒ Ty = w + 2 warm D # $ ! " L 1 ⇒ T = w + 2 warm csc θ D
these are given!
Translational Equilibrium !
Fx = 0
⇒ T x − Ex = 0 ⇒ Ex = Tx = T cos θ # $ ! " L 1 ⇒ Ex = w + 2 warm cot θ D ! Fy = 0 ⇒ Ty − w − warm − Ey = 0 ⇒ Ey = Ty + w + warm ! " ! " L L ⇒ Ey = w 1 + + warm 1 + D 2D
Rotational Equilibrium
!2 N !1 N
This stationary car is in rotational equilibrium because the torques ! 1, and N ! 2 cancel. due to w !, N
Rotational Equilibrium: Another Way
This stationary car is in rotational equilibrium because its centre of gravity is between the support points. No need to calculate torques! Just examine the geometry.
Rotational Equilibrium: Another Way
This stationary car is not in rotational equilibrium because its centre of gravity is not between the support points.
Rotational Equilibrium: Another Way Original car once more: This stationary car is in rotational equilibrium because its centre of gravity is between the support points.
Rotational Equilibrium: Another Way
This stationary truck is not in rotational equilibrium because its centre of gravity is not between the support points. Road angle is same as for car, but the truck’s cg is higher!
Kinetic Energy K= =
v
2 1 mv 2
m
2 1 m (ωr) 2
r =
! 1
=
2 1 2 Iω
2
mr
" 2
ω
2
where I is the “moment of inertia” 2
(In this case, I = mr )
Kinetic Energy v1 ¶ All objects have same ω r1
v3 v2
r3
r2 K=
! 1 2
" 2 2 2 2 1 m1 r1 + m2 r2 + ... ω = 2 Iω
where I is the “moment of inertia” ! " 2 2 2 (In this case, I = m1 r1 + m2 r2 + m3 r3 + ...
Energy Conservation
(before)
Energy Conservation
We can’t apply energy conservation yet, because we don’t know what to use for the moment of inertia in the pulley’s kinetic energy 2 .... 1 K = 2 Iω (after)
Moments of Inertia
I = MR
2
I=M
!
2 R1
+ 2
2 R2
"
I=
2 1 M R 2
Moments of Inertia
I=
2 1 M R 2
I=
2 1 M L 12
Moments of Inertia See Table 9.1 in your textbook
I=
2 2 M R 5
I=
2 2 M R 3
Moments of Inertia See Table 9.1 in your textbook
I=
2 2 M R 5
I=
2 2 M R 3
Energy Conservation Ui = mgh + M gH Ki = 0
+ 0
for block for pulley
y=0
(before)
Energy Conservation Uf = 0
v ω= R
+ M gH
Kf = 12 mv 2 + 12 Iω 2
I = 21 M R2
for block for pulley
(after)
Energy Conservation Uf = 0
v ω= R
+ M gH
Kf = 12 mv 2 + 12 Iω 2
I = 21 M R2
for block for pulley Ui + Ki = Uf + Kf
mgh + M gH = M gH + mgh = 12 mv 2 +
1 2 2 mv + 2 1 2 M v 4
2 " 2 v 2MR R2
! 1 1
(after)
Energy Conservation
v=
!
2mgh 1 m + 2M
Ui + Ki = Uf + Kf mgh + M gH = M gH + mgh = 12 mv 2 +
1 2 2 mv + 2 1 2 M v 4
2 " 2 v 2MR R2
! 1 1
(after)
Relations between Linear and Angular Measures (reminder) Rolling motion: • object spins about an axis through its centre of mass • the centre of mass moves along a line
If object rolls without slipping, then xcm = rθ, vcm = rω, acm = rα
Rolling down an Incline
For a single object at start, Ui = mgh and Ki = 0
Rolling down an Incline
due to translational motion
At end, Uf = 0 and Kf =
1 2 mv 2
+
1 2 Iω 2
due to spinning motion
Rolling down an Incline
v= Ui + Ki = Uf + Kf mgh =
1 2 2 mv
+
1 2 (f
2
· mR )
!
! v "2
2gh 1+f
R depends on the shape
Torque and Angular Acceleration τ = Iα
(This is the rotational analogue of F = ma in one-dimensional linear motion.)
Torque and Angular Momentum τ = Iα = I
dω dt
=
d dt
(Iω)
this is “angular momentum” If τ = 0 then Iω = constant
Ladder length = 5 m Sir L is 1/3 of the way up
No friction at wall !
Torque again ... Sir Lancelot on a ladder (or Example 9-4 in text): Free-body diagram of 5.0-m long ladder:
Rotation axis
Torque again ... Sir Lancelot on a ladder (or Example 9-4 in text): Ladder’s weight: τa = −(180 N)(1.5 m) = −270 N · m Sir Lancelot’s normal force: τb = −(800 N)(1.0 m) = −800 N · m Ground’s normal force: τ2 = +(n2 )(0.0 m) = 0 N · m Ground’s friction: τf = +(fs )(0.0 m) = 0 N · m Wall’s normal force: τ1 = +(n1 )(4.0 m) = 4n1
Torque again ... Sir Lancelot on a ladder (or Example 9-4 in text): Rotational equilibrium ! ⇒ τ =0 ⇒ τa + τb + τ2 + τf + τ1 = 0 ⇒ 4n1 = (270 + 800) N · m ⇒ n1 = 267.5 N
Translational equilibrium !
⇒ Fx = 0 ⇒ fs = n1 = 267.5 N ! and Fy = 0 ⇒ n2 = wa + wb = 980 N
Rotational Equilibrium ~ Building a mobile:
Find this Use
!
τ =0
– but first, choose a rotation axis!
Rotational Equilibrium ~ Building a mobile:
Rotation axis τm = +(mg)(0.10 m) τa = −(0.10g N)(0.15 m) τb = −(0.10g N)(0.40 m) τT = +(T )(0.0 m)
!
τ =0 ⇒ τm + τa + τb + τT = 0 ⇒ (mg)(0.1) = (0.1g)(0.55) ⇒ m = 0.55 kg
Rotational Equilibrium w , warm , θ , and all distances are given. ( θ is the angle between T and the horizontal)
Find T , Ex and Ey .
Rotational Equilibrium Rotation axis
Rotational Equilibrium Dumbbell: τd = +(w)(L) Forearm’s weight: τa = +(warm )(L/2) Horizontal component of T: τx = +(Tx )(0) = 0 Vertical component of T: τy = −(Ty )(D) Horizontal component of E: τh = +(Ex )(0) = 0 Vertical component of E: τv = +(Ey )(0) = 0
Rotational Equilibrium !
τ =0 L warm 2
⇒ wL + − Ty D# = $ 0 ! " L 1 ⇒ Ty = w + 2 warm D # $ ! " L 1 ⇒ T = w + 2 warm csc θ D
these are given!
Translational Equilibrium !
Fx = 0
⇒ T x − Ex = 0 ⇒ Ex = Tx = T cos θ # $ ! " L 1 ⇒ Ex = w + 2 warm cot θ D ! Fy = 0 ⇒ Ty − w − warm − Ey = 0 ⇒ Ey = Ty + w + warm ! " ! " L L ⇒ Ey = w 1 + + warm 1 + D 2D
Rotational Equilibrium
!2 N !1 N
This stationary car is in rotational equilibrium because the torques ! 1, and N ! 2 cancel. due to w !, N
Rotational Equilibrium: Another Way
This stationary car is in rotational equilibrium because its centre of gravity is between the support points. No need to calculate torques! Just examine the geometry.
Rotational Equilibrium: Another Way
This stationary car is not in rotational equilibrium because its centre of gravity is not between the support points.
Rotational Equilibrium: Another Way Original car once more: This stationary car is in rotational equilibrium because its centre of gravity is between the support points.
Rotational Equilibrium: Another Way
This stationary truck is not in rotational equilibrium because its centre of gravity is not between the support points. Road angle is same as for car, but the truck’s cg is higher!
Kinetic Energy K= =
v
2 1 mv 2
m
2 1 m (ωr) 2
r =
! 1
=
2 1 2 Iω
2
mr
" 2
ω
2
where I is the “moment of inertia” 2
(In this case, I = mr )
Kinetic Energy v1 ¶ All objects have same ω r1
v3 v2
r3
r2 K=
! 1 2
" 2 2 2 2 1 m1 r1 + m2 r2 + ... ω = 2 Iω
where I is the “moment of inertia” ! " 2 2 2 (In this case, I = m1 r1 + m2 r2 + m3 r3 + ...
Energy Conservation
(before)
Energy Conservation
We can’t apply energy conservation yet, because we don’t know what to use for the moment of inertia in the pulley’s kinetic energy 2 .... 1 K = 2 Iω (after)
Moments of Inertia
I = MR
2
I=M
!
2 R1
+ 2
2 R2
"
I=
2 1 M R 2
Moments of Inertia
I=
2 1 M R 2
I=
2 1 M L 12
Moments of Inertia See Table 9.1 in your textbook
I=
2 2 M R 5
I=
2 2 M R 3
Moments of Inertia See Table 9.1 in your textbook
I=
2 2 M R 5
I=
2 2 M R 3
Energy Conservation Ui = mgh + M gH Ki = 0
+ 0
for block for pulley
y=0
(before)
Energy Conservation Uf = 0
v ω= R
+ M gH
Kf = 12 mv 2 + 12 Iω 2
I = 21 M R2
for block for pulley
(after)
Energy Conservation Uf = 0
v ω= R
+ M gH
Kf = 12 mv 2 + 12 Iω 2
I = 21 M R2
for block for pulley Ui + Ki = Uf + Kf
mgh + M gH = M gH + mgh = 12 mv 2 +
1 2 2 mv + 2 1 2 M v 4
2 " 2 v 2MR R2
! 1 1
(after)
Energy Conservation
v=
!
2mgh 1 m + 2M
Ui + Ki = Uf + Kf mgh + M gH = M gH + mgh = 12 mv 2 +
1 2 2 mv + 2 1 2 M v 4
2 " 2 v 2MR R2
! 1 1
(after)
Relations between Linear and Angular Measures (reminder) Rolling motion: • object spins about an axis through its centre of mass • the centre of mass moves along a line
If object rolls without slipping, then xcm = rθ, vcm = rω, acm = rα
Rolling down an Incline
For a single object at start, Ui = mgh and Ki = 0
Rolling down an Incline
due to translational motion
At end, Uf = 0 and Kf =
1 2 mv 2
+
1 2 Iω 2
due to spinning motion
Rolling down an Incline
v= Ui + Ki = Uf + Kf mgh =
1 2 2 mv
+
1 2 (f
2
· mR )
!
! v "2
2gh 1+f
R depends on the shape
Torque and Angular Acceleration τ = Iα
(This is the rotational analogue of F = ma in one-dimensional linear motion.)
Torque and Angular Momentum τ = Iα = I
dω dt
=
d dt
(Iω)
this is “angular momentum” If τ = 0 then Iω = constant
