The rotor (flywheel) of a toy gyroscope has mass 0.150 kg. Is moment of inertia about its axis is 1.20 104kg2. The mass of the frame is 2.70 102kg. The gyroscope is supported on a single pivot with its center of mass a horizontal
Answers
Answer:
force exerted by the pivot is
M is the mass of the fly wheel , ? is the precessionangular speed
r is the horizontal distance
for angular speed we can apply the formula
W is the weight = m g= 0.0250 kg * 9.8 m/s
2
ris the horizontal distance = 0.04 m
I is the moment of inertia =
? is the precession angular speed = 1 rev/ 2.20s= 2? rad /2.20 s = 2.85 rad/s
you plug the values andismplify you will get the answer
The given question is incomplete and incorrect. Refer to the question given below:
The rotor (flywheel) of a toy gyroscope has a mass 0.150 kg. Its moment of inertia about its axis is 1.30 × 10⁴ kgm². The mass of the frame is 0.0255 kg. The gyroscope is supported on a single pivot with its centre of mass at a horizontal distance of 4.05 cm from the pivot. The gyroscope is precessing in a horizontal plane at the rate of 1 revolution in 2.3 seconds.
Find i) Find the upward force exerted by the pivot.
ii) Find the angular speed with which the rotor is spinning about its axis. (rev/min)
Concept:
To determine the force, the total mass of the gyroscope needs to be calculated and then we need to apply the force equation.
Given:
Mass of rotor = 0.150 kg
Moment of inertia = 1.30 × 10⁴ kg.m²
The mass of the frame is 0.0255 kg
Centre of mass distance = 4.05 cm = 0.0405 m
Time = 2.3 seconds
Find:
We need to determine:
i) the upward force exerted by the pivot.
ii) the angular speed with which the rotor is spinning about its axis.
Solution:
i) First we need to calculate the total mass of the gyroscope
It can be calculated as- Total mass of gyroscope = Mass of rotor + mass of the frame
The total mass of gyroscope = 0.150 + 0.0255
The total mass of gyroscope = 0.1755 kg
We know, Force = mass × acceleration due to gravity
a = g = 9.8
Therefore, Force = Total mass of gyroscope × acceleration due to gravity
Force = 0.1755 × 9.8
Force = 1.72 N
ii) To determine the angular speed-
The equation of angular speed is given as Angular speed = wr/i × 2π/t
angular speed = Force × Centre of mass distance / Moment of inertia × 2π/ time
Angular speed = 1.72 × 0.0405/ 1.3 × 10⁴ × 2π/2.3
Angular speed = 196.15 rad/s
In terms of revolution, it becomes-
1 rad/s = 9.5493 rev/min
Therefore, 196.15 rad/s = 196.15 × 9.5493 rev/min
= 1.873 × 10³ rev/min
Thus, i) the upward force exerted by the pivot is 1.72N
ii) the angular speed with which the rotor is spinning about its axis is 1.873 × 10³ rev/min.
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