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3 years ago

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An electric fan has a low speed of 3.00 rotations per second and a medium speed of 9.00 rotations per second. The fan rotates co

unter-clockwise, CCW. a) The fan is running on low speed when it is switched to medium speed. It takes 4.00 seconds for the fan to reach medium speed. What is the angular acceleration, LaTeX: \alphaα? b) If the rotational inertia of the fan is Ifan = 1.50 x 10-3 kg-m2, calculate the Net torque, LaTeX: \tau_{net}τ n e t, required for part a.

1 answer:

Korolek [52]3 years ago

7 0

Answer:

α=9.42 rad/s T= 0.014 rad/s $^{2}$

Explanation:

Low speed= Wi=3 rps = 3*2*π rad/s= 6π rad/sec

Medium speed= Wf= 9 rps= 9*2*π rad/sec= 18 rad/sec

time= t= 4 sec

Angular acceleration=α= $\frac{Wf-Wi}{t}$

⇒α=(18 π - 6 π)/4

⇒α=9.42 rad/s

b).

Rotational inertia= I = 1.5*10^-3 kg-m^2

Net torque= T = I*α = (1.5*10^-3) * (9.42) rad/s^2

⇒T= 0.014 rad/s^2

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Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acti

Tems11 [23]

Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

$\dfrac{dF_{net}}{df}=0$

We need to calculate the unknown force

Using formula of net force

$\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}$

Put the value into the formula

$\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}$

$\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}$

The magnitude of net force,

$F_{net}=\sqrt{F_{x}^2+F_{y}^2}$

$F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}$

$F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}$

$F_{net}=\sqrt{F^2+4899.78+36.232F}$

On differentiating w.r.to F

$(\dfrac{dF_{net}}{dF})^2=2F+36.232$

$0=2F+36.232$

$F=-\dfrac{36.232}{2}$

$F=-18.116\ lb$

Negative sign shows the direction of force which is downward.

Hence, The unknown force will be 18.116 lb.

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2 years ago

Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI)

Ludmilka [50]

Answer:

a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.

Explanation:

a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:

$a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$a' = 3.569$

The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:

$v_{f} = v_{o} + a \cdot t$

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v_{f}$ - Final speed, measured in meter per second.

$a$ - Acceleration, measured in $\frac{m}{s^{2}}$ .

$t$ - Time, measured in seconds.

$t = \frac{v_{f}-v_{o}}{a}$

$t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }$

$t = 0.095\,s$

Lastly, the severity index is now determined:

$SI = \frac{a'^{5}}{2\cdot t}$

$SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}$

$SI = 3047.749$

b) The initial and final speed of the injured are $1.944\,\frac{m}{s}$ and $5.278\,\frac{m}{s}$ , respectively. The travelled distance can be determined from this equation of motion:

$v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s$

Where $\Delta s$ is the travelled distance, measured in meters.

$\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}$

$\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}$

$\Delta s = 0.345\,m$ .

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