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REY [17]
3 years ago
6

Un objeto de 200 gramos está amarrado del extremo de una cuerda y gira describiendo un círculo horizontal de 1.20 m de radio a r

azón de 3 rev/seg la tensión en la cuerda es?
Physics
1 answer:
vesna_86 [32]3 years ago
5 0

Answer:

La tensión es 85.3 N.

Explanation:

Cuando el objeto gira en dirección horizontal, la sumatoria de fuerzas se puede calcular usando la segunda ley de Newton:  

\Sigma F_{x} =ma_{c}

T = ma_{c}  

Dado que el movimiento es horizontal, el peso (que está en el eje y) no contribuye en la sumatoria de fuerzas en el eje x. Por lo que la única fuerza actuando sobre el objeto en la dirección del movimiento es la tensión.  

En donde:                                          

m: es la masa del objeto = 200 g = 0.200 kg

a_{c}: es la aceleración centrípeta

La aceleración centrípeta viene dada por:  

a_{c} = \omega^{2} r

En donde:    

ω: es la velocidad angular del objeto = 3 rev/s

r: es el radio = 1.20 m

Entonces, la tensión es:

T = m\omega^{2} r = 0.200 kg(3\frac{rev}{s}*\frac{2\pi rad}{1 rev})^{2}*1.20 m = 85.3 N

   

Por lo tanto, la tensión es 85.3 N.  

Espero que te sea de utilidad!                                                                          

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<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to 90^\circ or \displaystyle \frac{\pi}{2} radians.

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Solve this equation for \omega(\text{final}) in terms of a_T and r:

\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}.

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\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}.

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\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608.

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