A gyroscope consists of a uniform disc of mass radius M= 1 kg and radius R= 0.1 m . The disc spins with an angular speed ω= 300 rad⋅s-1 as shown in the figure below. The gyroscope precesses, with its axle at an angle 30∘ below the horizontal (see figure). The gyroscope is pivoted about a point d= 0.4 m from the center of the disc. What is the magnitude of the precessional angular velocity Ω (in radians/sec)?

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To find the magnitude of the precessional angular velocity Ω, we can use the equation for the precession of a gyroscope:

Ω = (MgRcos(θ)) / (Idω)

Where:
M = mass of the disc
g = acceleration due to gravity
R = radius of the disc
θ = angle the axle makes with the horizontal
I = moment of inertia of the disc about its axis
d = distance from the center of the disc to the pivot point
ω = angular speed of the disc

First, let's calculate the moment of inertia I for the disc. The moment of inertia for a disc about its axis can be calculated using the formula:

I = (1/2)MR^2

Substituting the given values:
M = 1 kg
R = 0.1 m

I = (1/2)(1 kg)(0.1 m)^2
= 0.005 kg⋅m^2

Next, we need to calculate the angle θ in radians. Given that the axle is 30 degrees below the horizontal, we can convert it to radians:

θ = (30 degrees) x (π/180 degrees)
= 0.5236 radians

Now, let's substitute all the values into the equation for Ω:

Ω = (MgRcos(θ)) / (Idω)

Substituting the given values:
M = 1 kg
g = 9.8 m/s^2 (acceleration due to gravity)
R = 0.1 m
θ = 0.5236 radians
I = 0.005 kg⋅m^2
d = 0.4 m
ω = 300 rad/s

Ω = (1 kg)(9.8 m/s^2)(0.1 m)cos(0.5236 radians) / (0.005 kg⋅m^2)(0.4 m)(300 rad/s)
= (0.98)(0.866) / (0.005)(0.4)(300)
≈ 0.189 radians/sec

Therefore, the magnitude of the precessional angular velocity Ω is approximately 0.189 radians/sec.