Question

Consider the system shown in the figure below. The mass moment of inertia of the bar about the point O is JO, and the torsional stiffness of the spring attached to the pivot point is kt . Assume that there is gravity loading. The centre of gravity of the bar is midways, as shown in the figure.

Question 2 Consider the system shown in the figure below. The mass moment of inertia of the bar about the point O is Jo, and the torsional stiffness of the spring attached to the pivot point is k. Assume that there is gravity loading. The centre of gravity of the bar is midways, as shown in the figure. (a) Derive the equation of motion using Newton's second law (rotation) by summing moments about point O. (6) Linearize the equation of motion you found in (a) by using the 'small angle approximation', sin(O) = 0 (c) Find the potential energy of the system (NO small angle approximation) (d) Find the equivalent stiffness of the system assuming a small angle approximation cos(O) = 1 - - for energies, it is the quadratic terms that make all the action (motion) happen. So although we 'normally' use a small angle approximation of cos(0) = 1, when it comes to energy terms, you want to keep the quadratic term, otherwise we will get no motion. So in an energy term, we use cos(0) =1-for our small angle approximation for cosine. (e) What's the smallest value that k can be for any oscillations to happen (ie the assembly oscillates instead of tipping over)?

Answer 1

Let be the acceleration due to gravity.

(a)

Let be mass of the link.

= W = 7W+7V9 = 9

Free-body diagram:

_kᎾ Ꮎ Ꮇ g

► Mg sin e Mgcos e ktᎾ .

From the above figure, using summation of moments at point ,

Mg sin 0 - 0 = Joo

310 Joä + k0 = ogsin

[And from the figure, using summation of forces, the reaction can be found as , but this information is not asked in this question.]

R= Mg cose

(b)

If , then

sindo

3 Jo Joë+k40 = 300 g

--( )=0

[The above differential equation is the linearised equation of motion]

(c)

To calculate potential energy , the datum is assumed to be shown as in the figure below.

cos e Datum

P.E. = Mocos 0 = 3/gcos e

P.E. = 34 g cose

[Any reference can be assumed to be the datum, which would then remain to be the datum for the entire problem.]

(d)

If , then

-I = 6500

P.E. = g (1 - )

Equivalent stiffness:

Potential energy of a system of torsional stiffness $k_e$ is given by $-\frac{1}{2}k_e\theta^2$ .

By comparing the two expressions,

constant - k02 -1 ) 02

Therefore,

- ked? =-1 )

=ke = 3ung

[Constant is omitted, as the datum can always be adjusted to nullify the constant value of potential energy.]

(e)

The equation of motion is

--( )=0

For oscillations, let us assume that

0 = Asin (wt+0)

0 = - Aw2sin (wt+o) = -w20

+8+w20=0

By comparing with the above equation with the equation of motion, it can be observed that

$\omega^2=\frac{k_t}{J_0}-\frac{3g}{2L}$

For oscillations to happen,

>0

<T0<

Therefore, minimum value of $k_t$ for the system to oscillate is $\frac{3J_0g}{2L}$ .