Gyroscopic Couple and Rotational Motion: Detailed Solutions

This assignment solution details the calculations for gyroscopic couple and rotational motion problems. It includes finding the mass moment of inertia of a body about different axes, analyzing the gyroscopic couple acting on a spinning disk, and determining the change in kinetic energy of a wheel. Specific calculations involve determining mass (M1 and M2), moment of inertia (Ix1, Ix2, and Ix), magnitude of moment, and final velocity (V). The document also addresses the angular velocity of a disk after impact from a bullet, providing a comprehensive overview of the problem-solving process in mechanical engineering. Desklib provides a platform for students to access such solved assignments and past papers.

Solution

M1 = 7850 * 2 * 0.09*0.02 * 0.03
= 0.8478 kg
M2 = 7850∗π ( 0.02 )2
4 ∗0.05 = 0.1233 kg
Finding the mass of moment of inertia of 1 about x – axis
Ix1 = 1
12∗0.8478∗[ ( 2∗0.09 )2 + ( 0.03 )2 ]
= 0.002353 kgm2
Finding the mass of moment of inertia of 2 about x – axis
Ix2 = [ 1
2∗0.1233∗( 0.02
2 )2
]+ [ 0.1233∗ ( 0.09−0.03 )2 ]
= 0.00045 kgm2
Finding the mass moment of inertia of the entire body about x – axis
Ix = Ix1 + 2Ix2
= 0.002353 + 2(0.00045)
= 0.003253 kgm2
Finding the mass momentum of inertia of the entire body about x’ – axis
Ix’ = Ix + [(m1 + m2)(0.09 – 0.03)2]
= 0.003253 + [(0.85 + 2(0.12))(0.06)2]
= 0.007177 kgm2

In the spinning disk, due to ration of beam (support axial y (about y) and disk rotate with respect
to z axis then there must be acting gyroscopic couple which is the rotate of change of angular
momentum, and this gyroscopic couple tends to moment on the pin (A) which attached with disk
as same
ω=20 rad / sec, ωp = 6 rad/sec
Magnitude of moment = I∗ω∗ωp
m = 20 kg

r = disk radius
r = 150 mm
Magnitude of moment = 1
2 mr 2 ω∗ωp
= 1
2∗20∗0.152∗20∗6
= 27 Nm
Given mass = 15 kg

Wheel radius R = 200 mm = 0.2
Hub radius r = 50 mm = 0.05 m
Radius of gyration K = 130 mm = 0.13
after x = 3 m
Change in kinetic energy = work done (W)
W = mgsin θx
Change in kinetic energy = kf - kf
Wheel at rest ki = 0
∆ K . E= 1
2 I w2 + 1
2∗m∗v2 (Rotational + linear)
1
2∗m∗k2∗v2
r2 + 1
2∗m∗v2 = mgsinθ(x )
V2 = 2mgxsinθ∗r2
m(k 2+r2) = 2∗g∗sinθ∗r2
k2 +r 2
V = √ 2∗g∗sinθ∗r2
k2+ r2
V = √ 2∗9.81∗3∗sin 60
0.132 +0.052 ∗0.05
= 0.89042 m/s
V ≅ 0.9 m/ s

The mass moment of inertia of disk is
I = m R2
2 = 5∗0.22
2 = 0.1
The initial angular moment from point O is
Take the angle be 300
Li = mbullet * V * rcos300
rcos(30) is the perpendicular distance from O to along the line of action of bullet motion
Li = 7*10-3 * 693 * 0.2*cos300 = 0.84
Now the final moment is given by
Lf = Iω

0.1ω=0.84
The angular velocity of the disk is
ω=8.4 rad /s