[]= [] [] [] []= []= [] [] K ⋅ l m ⋅ l ⋅ t ⇒ K m ⋅ l ⋅ t [] []= []= [] [] [] []= K ⋅ l m ⋅ l ⋅ t ⇒ K m ⋅ t K ⋅ l m ⋅ l ⋅ t ⇒ K m ⋅ l ⋅ t r r r () ()= ∇ ∧ F = 0 ⇒∂ ∂ F F F F F ∂ F € () −∂ = 0, −∂ = 0, −∂ = 0 ∂ z ∂ x ∂ z ∂ y ∂ y ∂ x r r ()= F = ∇ UF ∂ U ∂ U

(1)

€

F =r r

∇ U F_x =∂U

∂x, F_y=∂U

∂y , F_z=∂U

∂z b )

c) a)

€

K₁

[ ]

^{⋅ l}

[ ]

⁰ = m ⋅ l ⋅ t

[

⁻²

]

^⇒

^{[ ]}

^K¹ = m ⋅ l ⋅ t

[

⁻²

]

K₂

[ ]

^{⋅ l}

[ ]

¹ = m ⋅ l ⋅ t

[

⁻²

]

^⇒

^{[ ]}

^K² ^{= m ⋅ t}

[

⁻²

]

K₃

[ ]

^{⋅ l}

[ ]

² = m ⋅ l ⋅ t

[

⁻²

]

^⇒

^{[ ]}

^K³ ^{= m ⋅ l}

[

⁻¹^{⋅ t}⁻²

]

€

∇ ∧r r F =r

0 ⇒

∂F_x

∂z −∂F_z

∂x = 0, ∂F_y

∂z −∂F_z

∂y = 0, ∂F_x

∂y −∂F_y

∂x = 0

€

U x, y,z

( )

^{= K}1x −1

2K₂y²+1 3K₃z³

€

U x, y,z

( )

⁼

^∫

^{F ⋅ d}^r ^r^{r =}

^∫ (

^K1ˆ i − K₂yˆ j + K₃z²k ˆ

) (

^dxˆi + dyˆ j + dz ˆ k

)

⁼

^∫ ⁽

^K¹^{dx − K}²^{ydy + K}³^z²^dz

⁾

€

U O

( )

^{= U 0,0,0}

( )

^{= 0}

U B

( )

^{= U 0,3,0}

( )

^{= −9J}

L_OB = U B

( )

^{− U O}

( )

^{= −9J}

d)

(2)

Problema n. 4

€

P − T − ma₀= ma TR = I ˙ ω

 



Forze che agiscono sullo yo-yo:

€

F r _R = r F _A + r

F _T + r F _C

Equazioni cardinali della dinamica:

€

F r _A = r P + r

T

€

F r _T = −ma r ₀

€

F r _C = 0

€

ω =˙ a r R = a

R

€

TR = I a R T =1

2mR² a R=1

2ma

m g −/ 1

2m a − / / m a₀= / m a 3

2a = g − a₀ ⇒ a =2

3

(

g − a₀

)

^{⇒ a =}²

3

(

9.8 −1.5

)

^{= 5.53 ms}⁻²

(3)

Conservazione della componente orizzontale della quantità di moto:

€

Q_i = MV_i+ mv_i = 0

€

T_i=1

2MV_i²+1

2mv_i²= 0

Conservazione dell’energia meccanica:

€

ΔE = E_f − E_i= 0 ⇒ E_f = E_i ⇒ T_f + U_f = T_i+ U_i

€

Q_f = MV_f + mv_f = MV_A + mv_A

€

U_i= Mgh_M + mgh_m = Mg ⋅ 0 + mgh = mgh

€

T_f =1

2MV_f²+1

2mv²_f =1

2MV_A²+1 2mv_A²

€

U_f = Mg⋅ 0 + mg ⋅ 0 = 0

€

ΔQ = Q_f − Q_i= 0 ⇒ Q_f = Q_i

1

2MV_A² +1

2mv_A² = mgh

€

MV_A + mv_A = 0 ⇒ v_A = −M mV_A a)

(4)

€

1

2MV_A² +1 2mM²

m² V_A² = mgh

€

1

2M 1+ M m



  

  V_A²= mgh

€

V_A² = 2m²gh

M M + m

( )

^{⇒ V}^A ⁼

2m²gh M M + m

( )

⁼

2⋅1²⋅ 9.8⋅ 4

3 3+ 1

( )

^{= 2.56ms}

−1

€ b )

€

v_A = −M

mV_A = −3

12.56 = 7.67ms⁻¹

€

1

2mv_U² =1

2mv_A² − L_T_r

€

T r _r= −µr

N = −µmg

€

L_T_r =µmg dx

0 d

∫

⁼^µmgx⁰^d ⁼^µmgd

€

1

2m v/ _U² =1

2m v/ _A² −µ / m gd

€

v_U² = v_A² − 2µgd

€

v_U = v_A² − 2µgd = 58.80 − 2⋅ 0.5 ⋅ 9.8⋅ 5 = 9.80 = 3.13ms⁻¹ 1

2m v/ _U² = / m gh_M h_M =v_U²

2g = 9.80

2⋅ 9.8= 0.5m

(5)

c)

€

t = t_d + t_p= t_d + ^p 4

€

µg = a_media =v_A − v_U t_d

€

t_d =v_A − v_U

µg = 7.67 − 3.13

0.5⋅ 9.8 = 0.93s

€

T_p = 2π l

g = 2π 8

9.8 = 5.68s

€

t = 0.93 +5.68

4 = 2.35s