Question

Motor at O is used to produce a moment M to rotate a mechanism shown in Figure Q4 in the vertical plane at a constant rate of θ 10 rad/s in the counter clockwise direction. The mechanism consists of a 1.5-kg small sphere which is rigidly welded to the bent arm OAB of negligible mass. a) Draw the free-body diagram and kinetic diagram of the mechanism. State the necessary assumptions. CO1/PO1/C2] (5 marks) b) Find the dynamic-force reaction at the support O and the moment applied by the motor [CO2/PO2/C4] (13 marks) c) Evaluate the moment applied by the motor if the mass of bent arm OAB is to be considered. CO3/PO3/C6] (2 marks) OA: 30 cm AB:40 cm Figure Q4

Answer 1

a) The FBD and kinetic diagrams are as follows:

$\dpi{100} \small \alpha$ is the angle the line joining O and B makes with the horizontal.

The assumptions are:

1. The mass of the connecting rod is negligible.

2. The center of mass of the ball lies exactly at B.

b) From the FBD and the kinetic diagram:

$\dpi{100} \small R_y-m_Bg=-m_B\theta^2l_{OB}\sin\alpha$

$\dpi{100} \small R_y=1.5\times9.81-1.5\times10^2\times(\sqrt{0.3^2+0.4^2}\,)\times\sin(\theta t+\tan^{-1}(l_{OA}/l_{AB}))$

$\dpi{100} \small R_y=14.715-75\sin(\theta t+36.87^\circ)$

Similarly,

$\dpi{100} \small R_x=-m_B\theta^2l_{OB}\cos\alpha$

$\dpi{100} \small R_x=-1.5\times10^2\times(0.5)\times\cos(\theta t+36.87^\circ)$

$\dpi{100} \small R_x=-75\cos(\theta t+36.87^\circ)$

And,

$\dpi{100} \small M-m_Bgl_{OB}\cos\alpha-(m_{OA}+m_{OB})x_{COM}=0$

$\dpi{100} \small M=1.5\times9.81\times0.5\cos(\theta t+36.87^\circ)$

$\dpi{100} \small M=7.3575\cos(\theta t+36.87^\circ)$

c) Masses of the rods are not given. Hence I'll keep the final answer in variables.

The center of mass of the individual rods is at the midpoints of the rods. The center of mass of both sections together is at the weighted mean of these coordinates:

$\dpi{100} \small x_{COM}=\frac{m_{OA}\times0.5 l_{OA}\cos(\alpha+\tan^{-1}(l_{AB}/l_{OA}))+m_OB\times( l_{OA}\cos(\alpha+\tan^{-1}(l_{AB}/l_{OA}))+0.5l_{AB}\cos(36.87-\alpha))}{m_{OA}+m_{OB}}$

$\dpi{100} \small x_{COM}=\frac{m_{OA}\times0.15\cos(\theta t+90^\circ)+m_{OB}\times(0.3\cos(\theta t+90^\circ)+0.2\cos(-\theta t))}{m_{OA}+m_{OB}}$

Hence the moment equation is,

$\dpi{100} \small M-m_Bgl_{OB}\cos\alpha-(m_{OA}+m_{OB})gx_{COM}=0$

$\dpi{100} \small M=0.5m_Bg\cos\alpha+0.15m_{OA}g\cos(\theta t+90^\circ)+m_{OB}g(0.3\cos(\theta t+90^\circ)+0.2\cos(\theta t))$

Note: For any clarifications, especially for part c, contact me in the comments.