Question

A gyroscope, a uniform disc of radius R=5.0cm at the end of a rod of length l=10cm (figure shown above), is mounted on the floor of an elevator car going up with a constant acceleration \[2m/{{s}^{2}}\]. The other end of the rod is hinged at the point O . The gyroscope precesses with an angular velocity n=0.5rps. Neglecting the friction and the mass of the rod, the proper angular velocity of the disc in rad/s is 100x. Find the value of x.

Answer 1

Hint: Here we can exploit the symmetry since the disc is uniform and it is rotating around its axis passing through its centre of mass. Also, the system is moving up so we have to adjust the value of new acceleration according to the problem.

Complete step by step answer:
The moment of inertia of the disc about an axis passing through its centre of mass is given by \[I=\dfrac{M{{r}^{2}}}{2}\]
Where M is the mass of the disc and r is its radius. If the angular velocity of the disc is ω, then its angular momentum can be written as,
\[\begin{align}
  & L=I\omega \\
 & L=\dfrac{M{{r}^{2}}\omega }{2} \\
\end{align}\]
The precession frequency is given as, \[\omega =2\pi n\]
Now we know since the body is rotating, torque is acting on it.
\[\left| \dfrac{d\overrightarrow{M}}{dt} \right|=\dfrac{M{{r}^{2}}\omega }{2}\times 2\pi n\], this must be equal to net torque acting on the body, so
\[\dfrac{M{{r}^{2}}\omega }{2}\times 2\pi n\]=\[M(g+a)l\]
\[\omega =\dfrac{(g+a)l}{\pi n{{r}^{2}}}\]--------------(1)
In equation (1) a is the acceleration of the elevator going upwards
g is the acceleration due to gravity
given values:
r=5.0cm= 0.05m
l=10cm= 0.1m
g=9.8 \[m/{{s}^{2}}\]
a= 2 \[m/{{s}^{2}}\]
n=0.5
on substituting the values in eq (1) we get
\[\omega \]= 300rad/s
But given in the question that the proper angular velocity of the disc in rad/s is 100x
Therefore, 300=100x
Which gives $ x=3. $

Note:
While substituting the values all the units must be in the same notation preferably in SI. If the frequency is given in round per minute, always convert it into round per second and if we multiply it by 2 \[\pi \] we get the angular frequency.
Also, torque is a vector quantity and is given by the cross product of force and perpendicular distance from the axis of rotation. In this problem both the vectors were perpendicular to each other, we have to see these all the things.