Question

Two children A and B of same mass (including their caps) M are sitting on a see – saw as chown in figure. Initially, the beam is horizontal. At Once, child B throws away his cap (mass M/25) which falls at point Q, midpoint of the left half of the beam, due to this the balance of beam is disturbed. To balance it again what is the mass m required to be put at point P on the right half of the beam?

• A. 0.16M
• B. 0.12M
• C. 0.04 M
• D. 0.1M

Answer 1

The correct option is B 0.12M

Initially, the beam was horizontal because the centre of mass of the system was at the centre (pivot of see-saw) of the system. When child B threw his cap to point Q, centre of mass of the system shifted to the left of the pivot and that’s why the balance got disturbed and beam started rotating in anticlockwise sense. To balance it again, we must put some mass to the right of it so as to displace the centre of mass of the system again at the centre of the system.
If a mass m is put at point P, to bring the centre of mass of the system at the centre, we have
$M\frac{L}{2}+\frac{M}{25}\frac{L}{4}=\frac{24M}{25}\frac{L}{2}+m\frac{L}{4}$
or m = 0.12 M