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3 years ago

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3. El tambor de una lavadora que gira a 3 000 revoluciones por minuto (rpm) se acelera uniformemente hasta que alcanza las 6 000

rpm, completando un total de 12 revoluciones.
d. Determina la aceleración tangencial, centrípeta y la total en m.s-2 cuando el tambor a alcanzado los 60000 rpm

e. Explica lo que ocurre con la magnitud y dirección de los vectores aceleración tangencial, aceleración centrípeta, aceleración total, aceleración angular, velocidad angular cuando la lavadora ha girado desde 3000 rpm hasta 6000 rpm.

1 answer:

cestrela7 [59]3 years ago

8 0

Answer:

d) α = 1693.5 rad / s² , a = 392.7 m / s² , a_total = α √(R² +1) ,

e) tan θ = a / α

Explanation:

This is an exercise in linear and angular kinematics.

We initialize reduction of all the magnitudes to the SI system

w₀ = 3000 rev / min (2π rad / 1rev) (1min / 60s) = 314.16 rad / s

w = 6000 rev / mi = 628.32 rad / s

θ = 12 rev = 12 rev (2π rad / 1 rev) = 75.398 rad

d) ask for centripetal, tangential and total acceleration.

Let's start by looking for centripetal acceleration, let's use the formula

w² = w₀² + 2 α θ

α = (w²- w₀²) / 2θ

we calculate

α = (628.32²2 - 314.16²) / 2 75.398

α = 1693.5 rad / s²

the quantity is linear and angular are related

the linear or tangential acceleration is

a = α R

where R is the radius of the drum

a = 1693.5 R

Unfortunately you do not give the radius of the drum for a complete calculation, but suppose it is a washing machine drum R = 20 cm = 0.20 m

a = 1693.5 0.20

a = 392.7 m / s²

the total acceleration is

a_total = √(a² + α²)

a_total = √ (α² R² + α²)

a_total = α √(R² +1)

e) The centripetal acceleration is directed towards the center of the movement is radial and its magnitude is constant

Tangential acceleration is tangency to radius and its value varies proportionally radius

the total accelracicon is the result of the vector sum of the two accelerations and their directions given by trigonometry

tan θ = a / α

the angular velocity increases linearly when with centripetal acceleration

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$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

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