Challenging +1.8 This is a challenging Further Maths mechanics problem requiring energy methods, rotational dynamics with a moving constraint, and careful force analysis on a non-inertial reference frame. The multi-step nature (energy → forces in rotating frame → friction condition), combined with the need to handle moment of inertia for a system with distributed mass and the subtlety of analyzing forces on the ring in the rod's rotating frame, places this well above average difficulty but not at the extreme end for Further Maths.
A uniform rod \(AB\), of length \(2a\) and mass \(2m\), can rotate freely in a vertical plane about a smooth horizontal axis through \(A\). A small rough ring of mass \(m\) is threaded on the rod. The rod is held in a horizontal position with the ring at rest at the mid-point of the rod. The rod is released from rest. Using energy considerations, show that, until the ring slides,
$$a\dot{\theta}^2 = \frac{18}{11}g \sin \theta,$$
where \(\theta\) is the angle turned through by the rod. [3]
Show that, until the ring slides, the magnitudes of the friction force and normal contact force acting on the ring are \(\frac{20}{11}mg \sin \theta\) and \(\frac{2}{11}mg \cos \theta\) respectively. [6]
The coefficient of friction between the ring and the rod is \(\mu\). Find, in terms of \(\mu\), the value of \(\theta\) when the ring starts to slide. [2]
Find MI of rod and ring about A: I = ⅓ 2ma + 2ma + ma M1
Use conservation of energy: ½ I (dθ/dt) 2 = 3mga sin θ M1
Substitute I = (11/3) ma 2 : a (dθ/dt) 2 = (18/11) g sin θ A.G. A1
Resolve forces on ring along rod (ignore sign of F): F = mg sin θ + ma (dθ/dt) 2 M1
Substitute for dθ/dt: F = (29/11) mg sin θ A.G. A1
Resolve forces on ring normal to rod (ignore signs): R = mg cos θ – ma d 2θ /dt 2 M1
Find d 2θ /dt 2 by differentiating eqn above: a d 2θ /dt 2 = (9/11) g cos θ M1 A1
Combine to give magnitude of normal force: R = (2/11) mg cos θ A.G. A1
Relate µ to θ : µ = F/R = (29/2) tan θ M1
Answer
Marks
Find θ : θ = tan –1 (2µ /29) A1
(3)
(6)
Answer
Marks
(2)
[11]
Question 4:
4 | 2 2 2
Find MI of rod and ring about A: I = ⅓ 2ma + 2ma + ma M1
Use conservation of energy: ½ I (dθ/dt) 2 = 3mga sin θ M1
Substitute I = (11/3) ma 2 : a (dθ/dt) 2 = (18/11) g sin θ A.G. A1
Resolve forces on ring along rod (ignore sign of F): F = mg sin θ + ma (dθ/dt) 2 M1
Substitute for dθ/dt: F = (29/11) mg sin θ A.G. A1
Resolve forces on ring normal to rod (ignore signs): R = mg cos θ – ma d 2θ /dt 2 M1
Find d 2θ /dt 2 by differentiating eqn above: a d 2θ /dt 2 = (9/11) g cos θ M1 A1
Combine to give magnitude of normal force: R = (2/11) mg cos θ A.G. A1
Relate µ to θ : µ = F/R = (29/2) tan θ M1
Find θ : θ = tan –1 (2µ /29) A1 | (3)
(6)
(2) | [11]
A uniform rod $AB$, of length $2a$ and mass $2m$, can rotate freely in a vertical plane about a smooth horizontal axis through $A$. A small rough ring of mass $m$ is threaded on the rod. The rod is held in a horizontal position with the ring at rest at the mid-point of the rod. The rod is released from rest. Using energy considerations, show that, until the ring slides,
$$a\dot{\theta}^2 = \frac{18}{11}g \sin \theta,$$
where $\theta$ is the angle turned through by the rod. [3]
Show that, until the ring slides, the magnitudes of the friction force and normal contact force acting on the ring are $\frac{20}{11}mg \sin \theta$ and $\frac{2}{11}mg \cos \theta$ respectively. [6]
The coefficient of friction between the ring and the rod is $\mu$. Find, in terms of $\mu$, the value of $\theta$ when the ring starts to slide. [2]
\hfill \mbox{\textit{CAIE FP2 2009 Q4 [11]}}