Physics - W.E.P. Question with Solution | TestHub

Let the velocity of the point mass at the highest point A be $v_A$ and at the horizontal point C be $v_C$.

By the principle of conservation of mechanical energy between points A and C (taking the horizontal level of C as the reference for zero potential energy):

Total Energy at A = Total Energy at C

$\frac{1}{2}m{v}_A^2+mg\ell =\frac{1}{2}m{v}_C^2$

Rearranging the above equation, we get:

$v_C^2-v_A^2=2g\ell$

The centripetal acceleration of a particle moving in a circle of radius $\ell$ with velocity $v$ is given by $a_c=\frac{v^2}{\ell }$.

The increase in centripetal acceleration when the mass moves from A to C is:

$\Delta a_c=a_{cC}-a_{cA}=\frac{v_C^2}{\ell }-\frac{v_A^2}{\ell }$

$\Delta a_c=\frac{v_C^2-v_A^2}{\ell }$

Substituting $v_C^2-v_A^2=2g\ell$ into the equation:

$\Delta a_c=\frac{2g\ell }{\ell }=2g$

Answer: $2g$