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2 years ago

¿Cuál es la aceleración centrípeta de un móvil que recorre una

Physics

1 answer:

Lady bird [3.3K]2 years ago

5 0

Answer:

a) a = 4.57 m/s², b) a = 6.48 m / s² , c) a = 1.42 m / s²,d) r = 82.3 m

Explanation:

The centripetal acceleration is the acceleration responsible for the change of direction of the acceleration vector and occurs in circular movements, the expression is

a = v² / r

let's apply this precaution to our cases

a) let's calculate

a = 8²/14

a = 4.57 m/s²

b) an automobile at v = 65 km / h (1000 m / 1km) (1 h / 3600 s) =18,055 m/s

let's reduce feet to meters

1 ft = 0.3048 m

r = 165 ft (0.3048 m / 1 ft) = 50.292 m

a = 18,055 2 / 50,292

a = 6.48 m / s²

c) we calculate

a = 1.25²2 / 1.1

a = 1.42 m / s²

d) we look for the radius

a = v² / r

r = v² / a

we reduce

v = 80 km / h (1000 m / 1km) (1h / 3600s) = 22.22 ms

r = 22.22²/6

r = 82.3 m

e) the cenripeta acceleration is used to take the curves on the highway,

Used in centrifuges to separate compounds

It is used in the games of the park of atraccio

Used in CD players and computer hard drives

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3 years ago

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

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Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

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$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

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3 years ago

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mars1129 [50]

Answer:

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Explanation:

therefore,

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