RBD BYU 27 - CHATUP707

Physics Problem 27 Solution

Fig 27. Free Body Diagram of Beads

Note: At the instant when it is released there is no angular velocity, so the components of acceleration of the beads along the rod must be same.

Since the rod is rigid, the relative acceleration along the rod is zero. Projecting the accelerations along the rod (at $45^\circ$):

$$ \frac{a_1}{\sqrt{2}} = \frac{a_2}{\sqrt{2}} $$ $$ a_1 = a_2 = a \quad (\text{Say}) $$

For the upper bead:
The equation of motion along the direction of acceleration (vertical):

$$ mg – \frac{N_1}{\sqrt{2}} = ma \quad \dots(1) $$ (Here, $N_1$ is the internal force from the rod, having a vertical component opposing gravity)

For the lower bead:
The equation of motion along the direction of acceleration (horizontal):

$$ \frac{N_1}{\sqrt{2}} = ma \quad \dots(2) $$

Solving for acceleration $a$:
Adding equation (1) and (2):

$$ (mg – \frac{N_1}{\sqrt{2}}) + (\frac{N_1}{\sqrt{2}}) = ma + ma $$ $$ mg = 2ma $$ $$ a = \frac{g}{2} $$

Calculating Normal Force $N$:
The normal force $N$ on the lower bead balances the downward forces (gravity + vertical component of rod force):

$$ N = mg + \frac{N_1}{\sqrt{2}} $$

Substituting $\frac{N_1}{\sqrt{2}} = ma = m(g/2)$ from equation (2):

$$ N = mg + \frac{mg}{2} $$ $$ N = \frac{3mg}{2} $$