Av ΣΜ = = Linear Kinematics v = Ax At Angular Kinematics Linear Kinetics Angular Kinetics ΔΕ 0-0₁ W = = t-ti ΣΕ = ma Ια At tf-ti Δω wf-wi α = = F-At= t-ti m(v₁ - v₁) At t-ti w = w₁ + at p = mv A0 = W = f.d ā = = At v = v₁ + at = (0 +700) t Ax = 2 1 Ax = vot+at² = (w + wo) t 2 1 A0 = wot+at² Ep = mg(Ah) = mv2 M At I(wf-w₁) = H = Iw W = M.0 M = Fd₁ Iw² Ek 2 2 kx² k02 v² = v² + 2a(Ax) w² = w² + 2α(AO) Ek Xi+1-xi-1 V₁ = s = re 2h x-12x1 + x{+1 a₁ = V₁₂ = rw h² (v₁ sin 0)² Yapex = yi a₁ = ra 2g 1 = Σ mur? Ee = 2 P = Fv Ee = 2 P = Mw I = ICM +md² [=1 1 vz ar == w²r Other Kinetics y-y₁ = (v₁sin )t, +gt Other Kinematics r |aTotal = a² + a² Hsegment Hlocal + Hremote = ICOMSWs + mr²wg = FFT

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During the last rotation, a hammer thrower maintains a constant acceleration of his hammer, from 28 m/s at the beginning of the rotation to 32 m/s at the time of release. a) If the radius of the hammer's rotations is 1.3 m, what is the total linear acceleration when the hammer has covered 3/4 of the last rotation?

1) Find the tangential acceleration of the hammer: m/s2

2) Find the tangential velocity after 3/4 of the final rotation: m/s. (Note, this is NOT just 3/4 of the way from 28 to 32 , because you travel further in a given time at higher speed.)

3) Find the radial acceleration of the hammer after 3/4 of the final rotation: m/s2.

4) Find the total acceleration after 3/4 of the final rotation: m/s2.

 

b) How far will the throw go, if the initial launch angle is 41 degrees above the horizontal, 1.9 m above the ground.

1) Find the time-of-flight: s

2) Find the horizontal velocity: m/s

3) Find the horizontal distance: m.

Av
ΣΜ
=
=
Linear Kinematics
v =
Ax
At
Angular Kinematics
Linear Kinetics
Angular Kinetics
ΔΕ
0-0₁
W =
=
t-ti
ΣΕ = ma
Ια
At
tf-ti
Δω
wf-wi
α =
=
F-At=
t-ti
m(v₁ - v₁)
At
t-ti
w = w₁ + at
p = mv
A0 =
W = f.d
ā = =
At
v = v₁ + at
= (0 +700) t
Ax =
2
1
Ax = vot+at²
= (w + wo) t
2
1
A0 = wot+at²
Ep = mg(Ah)
=
mv2
M At I(wf-w₁)
=
H = Iw
W = M.0
M = Fd₁
Iw²
Ek
2
2
kx²
k02
v² = v² + 2a(Ax)
w² = w² + 2α(AO)
Ek
Xi+1-xi-1
V₁ =
s = re
2h
x-12x1 + x{+1
a₁ =
V₁₂ = rw
h²
(v₁ sin 0)²
Yapex = yi
a₁ = ra
2g
1 = Σ mur?
Ee
=
2
P = Fv
Ee
=
2
P = Mw
I = ICM +md²
[=1
1
vz
ar == w²r
Other Kinetics
y-y₁ = (v₁sin )t, +gt
Other Kinematics
r
|aTotal = a² + a² Hsegment Hlocal + Hremote = ICOMSWs + mr²wg
=
FFT <HsFN
Xob = xoa + xab
FFT = μkFN
Хсом
=
CDApv²
Vdist Vprox + Vdist prox
FD=
e = -
2
1,2
-b±√b² - 4ac
2a
2mg
Vterm
Хсом
CDAp
2m2
m₁ +m₂ m₁ + m₂
Σ=1 mixi
m1-m2
+12
-Σ
Pixi
i=1
n
ΣΕΜ
v₁₂- v₁'
V₂-V1
ΣΕ
Li=1 mixi
Σ Li=1
n
=
=
hbounced
hdropped
Σα
i=1
Pixi
F
Mc
Material Properties
F = kx
σ =
=
Omax I
π(R4-4)
π(R4-4)
Tr
A = π(R² - ²)
1 =
J =
4
2
x
w
E =
bw
σ
1
1
= 1/2 bh³
VQ
Tmax =
Ib
Transcribed Image Text:Av ΣΜ = = Linear Kinematics v = Ax At Angular Kinematics Linear Kinetics Angular Kinetics ΔΕ 0-0₁ W = = t-ti ΣΕ = ma Ια At tf-ti Δω wf-wi α = = F-At= t-ti m(v₁ - v₁) At t-ti w = w₁ + at p = mv A0 = W = f.d ā = = At v = v₁ + at = (0 +700) t Ax = 2 1 Ax = vot+at² = (w + wo) t 2 1 A0 = wot+at² Ep = mg(Ah) = mv2 M At I(wf-w₁) = H = Iw W = M.0 M = Fd₁ Iw² Ek 2 2 kx² k02 v² = v² + 2a(Ax) w² = w² + 2α(AO) Ek Xi+1-xi-1 V₁ = s = re 2h x-12x1 + x{+1 a₁ = V₁₂ = rw h² (v₁ sin 0)² Yapex = yi a₁ = ra 2g 1 = Σ mur? Ee = 2 P = Fv Ee = 2 P = Mw I = ICM +md² [=1 1 vz ar == w²r Other Kinetics y-y₁ = (v₁sin )t, +gt Other Kinematics r |aTotal = a² + a² Hsegment Hlocal + Hremote = ICOMSWs + mr²wg = FFT <HsFN Xob = xoa + xab FFT = μkFN Хсом = CDApv² Vdist Vprox + Vdist prox FD= e = - 2 1,2 -b±√b² - 4ac 2a 2mg Vterm Хсом CDAp 2m2 m₁ +m₂ m₁ + m₂ Σ=1 mixi m1-m2 +12 -Σ Pixi i=1 n ΣΕΜ v₁₂- v₁' V₂-V1 ΣΕ Li=1 mixi Σ Li=1 n = = hbounced hdropped Σα i=1 Pixi F Mc Material Properties F = kx σ = = Omax I π(R4-4) π(R4-4) Tr A = π(R² - ²) 1 = J = 4 2 x w E = bw σ 1 1 = 1/2 bh³ VQ Tmax = Ib
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