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Dahasolnce [82]
3 years ago
6

A heavy solid disk rotating freely and slowed only by friction applied at its outer edge takes 120 seconds to come to a stop.

Physics
1 answer:
alisha [4.7K]3 years ago
5 0

Answer:

The time is 16 min.

Explanation:

Given that,

Time = 120 sec

We need to calculate the moment of inertia

Using formula of moment of inertia

I=\dfrac{1}{2}MR^2

If the disk had twice the radius and twice the mass

The new moment of inertia

I'=\dfrac{1}{2}\times2M\times(2R)^2

I'=8I

We know,

The torque is

\tau=F\times R

We need to calculate the initial rotation acceleration

Using formula of acceleration

\alpha=\dfrac{\tau}{I}

Put the value in to the formula

\alpha=\dfrac{F\times R}{\dfrac{1}{2}MR^2}

\alpha=\dfrac{2F}{MR}

We need to calculate the new rotation acceleration

Using formula of acceleration

\alpha'=\dfrac{\tau}{I'}

Put the value in to the formula

\alpha=\dfrac{F\times R}{8\times\dfrac{1}{2}MR^2}

\alpha=\dfrac{2F}{8MR}

\alpha=\dfrac{\alpha}{8}

Rotation speed is same.

We need to calculate the time

Using formula angular velocity

\Omega=\omega'

\alpha\time t=\alpha'\times t'

Put the value into the formula

\alpha\times120=\dfrac{\alpha}{8}\times t'

t'=960\ sec

t'=16\ min

Hence, The time is 16 min.

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Explanation:

For equilibrium, \sum M = 0.

So,   8 m \times mg - (10 m) T_{1} = 0

             T_{1} = \frac{8 \times mg}{10}

                        = \frac{8 \times 90 \times 9.8}{10}

                        = 705.6 N

Also, for equilibrium \sum F_{y} = 0

              T_{1} + T_{2} - mg = 0

or,         T_{2} = mg - T_{1}

                        = 90 \times 9.8 - 705.6

                        = 176.4 N

Thus, we can conclude that the tension in the first rope is 176.4 N.

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Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4
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Answer:

Período del tambor: T = 0.25\,s, fuerza sobre la prenda: F \approx 80.852\,N, velocidad lineal del tambor: v \approx 10.053\,\frac{m}{s}, velocidad angular del tambor: \omega \approx 25.133\,\frac{rad}{s}.

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante (\omega), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga (a), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor (T), en segundos:

T = \frac{1}{f} (1)

Donde f es la frecuencia, en hertz.

(f = 4\,hz)

T = \frac{1}{4\,hz}

T = 0.25\,s

Ahora determinamos la fuerza aplicada sobre la prenda (F), en newtons:

F = m\cdot a (2)

F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}} (2b)

Donde:

m - Masa de la prenda, en kilogramos.

r - Radio interior del tambor, en metros.

(m = 0.32\,kg, r = 0.4\,m, T = 0.25\,s)

F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}

F \approx 80.852\,N

La velocidad lineal de la lavadora es:

v = \frac{2\pi\cdot r}{T} (3)

(r = 0.4\,m, T = 0.25\,s)

v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}

v \approx 10.053\,\frac{m}{s}

Y la velocidad angular del tambor de la lavadora:

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

(T = 0.25\,s)

\omega = \frac{2\pi}{0.25\,s}

\omega \approx 25.133\,\frac{rad}{s}

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