The figure below shows a rigid assembly of a thin hoop (of mass m and radius R =

Question

The figure below shows a rigid assembly of a thin hoop (of mass m and radius R = 0.120 m) and a thin radial rod (of mass m and length L = 3.00 R). The assembly is upright, but if we give it a slight nudge, it will rotate around a horizontal axis in the plane of the rod and hoop, through the lower end of the rod. Assuming that the energy given to the assembly in such a nudge is negligible, what would be the assembly's angular speed about the rotation axis when it passes through the upside-down (inverted) orientation?

Explanation / Answer

Since the rod has length 3R and is said to be radial rather than diametrical, I will assume that one end is at the centre of the circle, and the other projects beyond the hoop.

'The assembly is upright..' I assume that this means that the assembly is in a vertical plane, that the rod is vertical, and that the projecting end is uppermost. None of this is stated, but the question makes sense (to me) if it is interpreted in this way.

We can solve the problem by calculating the energy of the system. In the original position the CoG of the rod has a height R above the axis, while in the lower position the CoG is R below the axis. The original PE of the system relative to the inverted position is therefore 3mgR.

The rotational KE of the system is 0.5*I*w^2 where I is the total moment of inertia (Ih + Ir) and w is the angular velocity. Equating KE to PE will allow us to find w.

For the hoop Ih = mR^2

for the rod Ir = (m*L^2)/3 = (9/3)*m*R^2 =3mR^2

I = 3m*R^2

(3/2)*m*R^2*w^2 = 3mgR

w^2 = 2g/R

So w=sqrt(2g/R)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.