Compute the force in the rod:
A rod ABC rotating at 20 rpm about a vertical axis through A, supports a ball of mass 10 kg at its lower end. It is fixed in position by the rod BD, by neglecting the masses of rods AC and BD. Compute the force F in the rod BD. Is the force tensile or compressive? Determine rpm when this force is equal to zero.
Solution
The free body diagram of the rod AC is illustrated in Figure (b). The ball may be considered as a particle moving in a horizontal circle of radius r = 1.5 sin 30^{o} = 0.75 m. As the ball is moving with constant angular velocity it will have centripetal force m v^{2}/r. We create the dynamic equilibrium by applying inertia force F_{i} = m v ^{2} / r acting radically outward from the path.
The magnitude of velocity = 2 π N × r / 60 (Here, this N is rpm)
∴ v = 2 π × 20 × 0.75/60 = 1.57 m / sec
∴ F_{1} = m v^{2} /r
=10 × 1.57^{2} /0.75
= 32.86 N
Supposing a force F in BD to be compressive it acts on AC as illustrated.
If we consider moments of all the forces acting on AC about A, components of reaction at A, A_{v} and A_{h} shall not have any effect and we obtain :
At point A,
∑ M _{A } = 0
F × 0.9 cos 30^{o} + 32.86 × 1.5 cos 30^{o} - 98.1 × 1.5 sin 30^{o} = 0
∴ F = 39.63 N
Since F is positive, it acts as illustrated and BD is in compression.
For F = 0, the velocity may be obtained by taking moment about A again.
At point A,
∑ M_{ A} = 0
98.1 × 1.5 sin 30^{o} - (m v^{2} /r )× 1.5 cos 30^{o} = 0
∴ v ^{2} = 98.1 × 0.75 × tan 30^{o} /10= 4.2478
v = 2.06 m / sec
v = 2 π r N /60
∴ 2.06 =( 2 π× 0.75/60 )× N
∴ N = 26.26 rpm