- Si deve progettare una curva inclinata in modo tale che un’automobile possa percorrerla alla velocità di 10 m/s senza fare affidamento sull’attrito tra le gomme e l’asfalto. Per via dei vincoli urbanistici, il raggio della curva deve essere di 24 m. Di quale angolo dovrà essere inclinata la curva che devono progettare?
N * cos(β) = Forza peso;
N * cos(β) = m g;
N = m g / cos(β);
N * sen(β) = Forza centripeta
N * sen(β) = m * v^2/R;
(m g /cosβ ) * sen(β) = m * v^2/R;
sen(β) / cos(β) = tan(β);
m * g * tan(β) = m * v^2/R; m si semplifica.tan(β) = v^2 / (g * R ) = 10^2 /(9,8 * 24) = 0,425β = tan^-1(0,425) = 23°.
- Si determini il modulo della velocità tangenziale di un punto solidale con la superficie terrestre, supponendo che la Terra sia una sfera di raggio
R terra =6380 km = 6,380 * 10^6 m.
– Si ricavi l’espressione dell’accelerazione centripeta in funzione della latitudine φ;
– si stimi il rapporto tra il modulo dell’accelerazione di gravità g e il modulo dell’accelerazione centripeta a. latitudine.
r = (Rterra) * cos φ.
v = 2 * 3,14 * r / T = ω * r;T = 24 ore = 86400 s;
ω = 2 * 3,14 / (86400 s) = 7,272 * 10^-5 rad/sr dipende dalla latitudine alfa: al polo angolo alfa = 90°; all’equatore alfa = 0°;
r = (Rterra) * cos(φ) = (6,380 * 10^6 m ) * cos(φ)
v = ω* r = (7,272 * 10^-5) * (6,380 * 10^6 * cos(φ);
accelerazione centripeta: a =ω^2 * (Rterra * cos(φ) ) ;a = (7,272 * 10^-5)^2 * 6,380 * 10^6 *cos(φ)
a = 0,034 * cos(φ)
all’equatore cos(0°) = 1
a = 0,034 m/s^2
al polo cos(90°) = 0
a = 0 m/s^2
g / a = 9,81 / (0,034 cos(φ ) = 288 / cos(φ)
g è 288 volte maggiore di a all’equatore dove a ha valore massimo
in quanto cos(90°) = 1.Il nostro peso è leggermente minore in quanto si ottiene facendo la differenza vettoriale fra il vero peso (in viola nella figura, la forza di gravità) e la forza centripeta (in rosso verso il centro). La forza centripeta è piccolissima, quindi non la percepiamo e non ci accorgiamo di essere in un sistema rotante, non inerziale.
3) Calcolare le grandezze caratteristiche del moto che la Terra compie intorno al Sole (il raggio medio dell’orbita è R= 1,5*10^8 km, il tempo che impiega a percorrere l’orbita è di T= 365 giorni, immaginiamo che il moto sia circolare e uniforme).
R = 1,5 * 10^11 m
T = 365 * 24 h * 60 min * 60 s = 3,154 * 10^7 s
velocità angolare: 2 pgreco/T
omega = 2 * 3,14 / T = 6,28 / 3,154 * 10^7 = 1,99 * 10^-7 rad/s
velocità tangenziale media : v = 2 pgreco R / T = omega * R
v = 1,99 * 10^-7 * 1,5 * 10^11 = 29900 m/s (circa 30000 m/s = 30 km/s);
accelerazione centripeta: a = v^2/R
a = 29900 ^2 / (1,5 * 10^11 ) = 6 * 10^-3 m/s^2 = 0,006 m/s^2 ;
(l’accelerazione è molto piccola, quindi non ci accorgiamo di viaggiare in curva, è come se viaggiassimo di moto rettilineo a velocità costante).
4) Perché la forza centripeta nel punto più alto del giro della morte deve essere come minimo uguale al peso? Perché la reazione vincolare deve essere maggiore uguale a 0?Nel moto circolare verticale esiste la forza centripeta Fc che è data dalla somma della forza peso Fp e della reazione vincolare Fr che può essere la tensione di una corda o la spinta di una pista.
Nel punto più basso della traiettoria: Fc = Fr – Fp; quindi la reazione vincolare Fr = Fc + Fp; nel punto più basso la reazione è molto alta e il corpo sembra pesare di più.Nel punto più alto le forze sono tutte verso il basso:Fc = Fr + Fp; quindi la reazione vincolare è Fr = Fc – Fp
Freazione diventa 0 quando Fcentripeta = Fpeso; se Fpeso > Fcentripeta allora il corpo cade.
m V^2/R > = mg
V > =√(g R); R è il raggio; g è l’accelerazione 9,8 m/s^2
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