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Mandarinka [93]
3 years ago
7

The Hubble Space Telescope is stabilized to within an angle of about 2-millionths of a degree by means of a series of gyroscopes

that spin at 19,200 rpm. Although the structure of these gyroscopes is actually quite complex, we can model each of the gyroscopes as a thin-walled cylinder of mass 2.0 kg and diameter 5.0 cm, spinning about its central axis. How large a torque would it take to cause these gyroscopes to precess through an angle of 1.0×10−6 degree during a 5.0-hour exposure of a galaxy?
Physics
1 answer:
Likurg_2 [28]3 years ago
7 0

Answer:

The answer to the question is;

The required torque that it would take to cause the gyroscopes to precess through an angle of 1.0×10−6 degree during a 5.0-hour exposure of a galaxy is 2.44 ×10⁻¹² N·m  

Explanation:

To solve the question we first resolve the units of the given quantities as follows

The gyroscopes spin at 19,200 rpm that is 19,200 revolutions per minute

1 revolution = 2π rad and

1 minute = 60 seconds

Therefore 19,200 revolutions per minute = 2π×19,200÷60 rad/s

= 2010.619 rad/s

The angle of precess is given as 1.0×10⁻⁶ °. We convert the angle to radians as follows

360 ° = 2π radians

1 ° = \frac{\pi }{180} radians and

1.0×10⁻⁶ ° =  \frac{\pi }{180} radians × 1.0×10⁻⁶ ° = 1.745×10⁻⁸ rad

To find the torque we note that the torque is given by

Precession angular speed × The moment of inertia × angular velocity

The precession angular speed is given by \frac{Precession. Angle}{time}

The precession angle was determined in rad as 1.745×10⁻⁸ rad

The precession time is 5 hours which is equal to 5×60×60 = 18000 s

Therefore the precession velocity = \frac{1.745*10^{-8} rad}{18000 s} =  9.696×10⁻¹³ rad/s

The moment of inertia is given by

Formula for the moment of inertia of a thin walled cylinder I = m·r²

Where:

r = Radius of the gyroscope = Diameter/2 = 5.0 cm/2 = 2.5 cm = ‪0.025‬ m

m = Mass of each gyroscopes = 2.0 kg

Therefore I = m·r² = 2.0 kg × (0.025‬ m)² = 0.00125 kg·m²

Torque, τ = Ω·I·ω

Where:

Ω = Precession velocity = 9.696×10⁻¹³ rad/s

I = Moment of inertia = 0.00125 kg·m²

ω = Angular speed = 2010.619 rad/s

τ = 9.696×10⁻¹³ rad/s × 0.00125 kg·m² × 2010.619 rad/s =

2.44 ×10⁻¹² kg·m²/ s² =   2.44 ×10⁻¹² N·m  

The required torque is 2.44 ×10⁻¹² N·m.

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7 0
3 years ago
In a women's 100-m race, accelerating uniformly, Laura takes 1.82 s and Healan 3.07 s to attain
Sav [38]

Answer:

Laura is ahead and for a distance of 3.22 m

Explanation:

To solve this problem of one-dimensional kinematics, we have to find the acceleration and the final speed of each runner. Let's start with Laura

Lura data is acceleration time 1.82s, total run time 10.4 s and total distance 100m.  In all the races the  rest starts, so the initial speed is zero (Vo = 0)

   Vf1= Vo + a1 t1    

   Vf1 = x/t                

   XT  = X1 + X2

   X1 = Vo t1 + ½ a1 t1²  

   X1 = ½ a1 t1²  

   X2 = Vf1 (t-t1)

This is the remaining time of the race after the acceleration is over.

    XT = ½ a1 t1² + Vf1 (t-t1)

We remplace the expression of Vf1

     XT = ½ a1 t1² + a1 t1 (t-t1)

Laura's aceleration (a1) is

   a1= XT / [ ½  t1² + t1 (t-t1)]

   a1= 100/ [ ½ 1.82²+ 1.82 (10.4 -1.82)]

   a1=  5.79m/s2  

We repeat the same calculation for the other Healan runner, whose data are: total distance 100m, acceleration time 3.07 s and total time 10.4 s

Vf2= Vo + a2 t2    

Vf2 = x/t                

XT  = X3 + X4

X3 = Vo t2 + ½ a2 t2²  

X3 = ½ a2 t2²  

X4 = Vf2 (t-t2)

XT = ½ a2 t2² + Vf2 (t- t2)

XT = ½ a2 t2² + a2 t2 (t-t2)

The aceleration of Healan (a2)

a2 = XT / [½ t2² + t2 (t-t2)]

a2 = 100 / [½ 3,07²+ 3.07 (10.4 -3.07)]

a2 = 3.67 m / s2

We also need the final speeds of each runner

Laura Vf1 = Vo + a1 t1

          Vf1 = 0 + 5.79 1.82

          Vf1 = 10.54 m / s

Healan Vf2 = Vo + a2 t2

            Vf2 = 0 + 3.67 3.07

            Vf2 = 11.27 m / s

Having the acceleration and speed of each runner, you can start answering the questions

a) For t3 = 6.15s

Laura

The time to stop with constant speed is what remains after accelerating

XL= ½ a1 t1² + Vf1 (t3-t1)

XL= ½ 5.79 1.82² + 10.54 (6.15 – 1.82)    

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Healan  

XH= ½ a2 t2² + Vf2 (t3-t2)

             XH= ½  3.67 3.07² + 11.27 (6.15-3.07)

             XH= 52.01 m

             (XL -XH)= 55.23- 52.01

             (XH -XL)=  3.22 m

It is appreciated from these results that Laura is ahead and for a distance of 3.22 m

b) If we analyze the acceleration values ​​of each runner, knowing that they leave the rest and that Healan at the end has a speed greater than Laura, the point of maximum distance difference is when Laura stops accelerating t = 1.82 s

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      XL= ½ 5.79 1.822

      XL= 9.59 m

      XH = ½ a2 t12

      XH= ½ 3.67 1.822

      XH= 6.08 m

The maximum distance difference is 3.51 m

c) Already analyzed in the previous part 1.82 s, since the Laura stop accelerating and Heala continue with acceleration will travel greater distances in equal time units

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