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3 years ago

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Given: The drag characteristics of a torpedo are to be studied in a water tunnel using a 1:5 scale model. The tunnel operates wi

th fresh water at 20°C, whereas the prototype torpedo is to be used in seawater at 15.6°C. To correctly simulate the behavior of the prototype moving with the velocity of 30 m/s, what velocity (in m/s) is required in the water tunnel?

1 answer:

Gnoma [55]3 years ago

4 0

Answer:

The velocity in the water tunnel is 128.717m/s

Explanation:

To find the result we need to Apply dynamic symiliraty, that is,

$\frac{V_m I_m}{\upsilon_m}= \frac{VI}{\upsilon}$

Where $\upsilon_i$ is the kinematic viscosity, V the velocity and I the lenght.

To find the kinematic viscosity it is necessary search in the table with this properties (I attached it)

To this kind of fluids, the properties to 15.6°c (Seawater) and 20°c (Water)are:

$\upsilon_m= 1.17*10^{-6}m^2/s$ (Seawater)

$\upsilon_m = 1.004*10^{-6}m^2/s$ (Water)

Replacing,

$\frac{V_m I_m}{\upsilon_m}= \frac{VI}{\upsilon}$

Solving to V_m

$V_m = \frac{I}{I_m}\frac{\upsilon_m}{\upsilon}V$

$V_m= \frac{5}{1} * \frac {1.004*10^{-6}}{1.17*10^{-6}} 30$

$V_m = 128.717m/s$

The velocity in the water tunnel is 128.717m/s

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FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

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