TURNTABLE SOLUTIONS
A gyroscope consists of a uniform disc of mass radius M= 1 kg and radius R= 0.3 m . The disc spins with an angular speed ω= 400 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)?
Ω=????
please help!!
Moment of inertial of the disc is I
I=0.5*M*R^2
Omega= (2*g*M*d/(I*angular speed)
Fill in the values and get the answer
it didn't work :(
maybe you made a mistake.. it works
NO it doesn't work because :
Omega=(m*g*d)/(I*angular speed)
without the '2'
:)
To find the magnitude of the precessional angular velocity Ω, we can use the following equation:
Ω = (ω * sin(θ)) / (d * cos(θ))
where:
- ω is the spinning angular speed of the gyroscope (given as 400 rad⋅s^-1),
- θ is the angle between the axle of the gyroscope and the horizontal (given as 30 degrees),
- d is the distance from the center of the disc to the pivot point (given as 0.4 m).
First, we need to convert the angle θ from degrees to radians:
θ = 30 degrees * (π/180 degrees) = 0.5236 radians
Now, we can substitute the given values into the equation:
Ω = (400 rad⋅s^-1 * sin(0.5236 radians)) / (0.4 m * cos(0.5236 radians))
Calculating the individual terms in the equation:
sin(0.5236 radians) ≈ 0.5
cos(0.5236 radians) ≈ 0.866
Ω = (400 rad⋅s^-1 * 0.5) / (0.4 m * 0.866)
Simplifying:
Ω ≈ 500 / 0.347
Calculating the final result:
Ω ≈ 1440.92 rad⋅s^-1
Therefore, the magnitude of the precessional angular velocity Ω is approximately 1440.92 radians/sec.