Question

Question 1. A uniform circular disc of mass m and radius r is pivoted at point 0, as shown in Figure 1. The disc is released from the position shown. Immediately after the release: (a) Obtain the angular acceleration of the disc in terms of r,1,g, and 0. [5 marks) [8 marks) Assuming r = 0.5 m and @ = 30° and using plot function in MATLAB: (b) Determine how the initial angular acceleration changes as I is varried from 0 to r, and discuss the results. (C) Find the optimum value of 1 for which the pendulum has maximum angular acceleration when released from = 30°. Does this value change if the release angle changes? Briefly explain why. [7 marks) The second moment of inertia of the disc around its centre of mass Gis: Total: [20 marks or G Figure 1

Answer 1

(a) The FBD of the disc is shown in the figure below.

Fx and Fy are the reaction forces at the pivot. Using the parallel axis theorem, the moment of inertia of the disc about the pivot can be computed as follows

$I_{p}=I_{G}+ml^{2}=m\left ( \frac{r^{2}}{2}+l^{2} \right )$ ...... (1)

Taking moment about the pivot and applying Newton's second law equivalent to rotational motion, we get

  

1 qö = mgsin (5-0) x1= mglcost

On substitution of I_p from equation (1), the initial angular acceleration can be expressed as

$\ddot{\theta }=\frac{2glcos\theta }{r^{2}+2l^{2}}$ ...... (2)

(b) The variation of angular acceleration with l can be plotted using equation (2). The MATLAB code and plot are as follows

>> r = 0.5; >> th = - pi/6; Theta >> 1 = 0:0.01:0.5; >> for i = 1:length(1) Acc(i) (9.81*r*l(i)*cos(th))/(r^2+(2*1(i)*2)); g = 9.81 m/s^2 end >> plot(1, Acc) >> |

3.5 X: 0.35 Y: 3.004 3 2.5 Angular acceleration 0.5 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 -

(c) The optimal value of l can be found by differentiating equation (2) with respect l and equating it to zero as follows

$\frac{\mathrm{d} \ddot{\theta }}{\mathrm{d} l}=\frac{\left ( r^{2}+2l^{2} \right )\times 2gcos\theta -2glcos\theta\times 4l }{\left ( r^{2}+2l^{2} \right )^{2}}=0$

On simplification we obtain

$l=\frac{r}{\sqrt{2}}$ ...... (3)

Equation (3) gives the optimal length 'l' for which the angular acceleration is maximum. Also this optimal length does not depend on the angle of release as the angle of release does not appear in equation (3) (it gets cancelled out). Kindly note that the magnitude of initial acceleration depends on angle of release but for any angle of release the maximum occurs at l given by equation (3). For parameters given in section b, substituting r = 0.5 in equation (3), we get l = 0.3536. This value of l at which maximum occurs matches that from the plot above.