1. Vertically oriented circular disks have strings wrapped around them. The other ends of the strings are attached to
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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 ° = radians and
1.0×10⁻⁶ ° = 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
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 = = 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.