A thin uniform rod of length L and certain mass is kept on a frictionless horizontal table with a massless string of length L fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point O . If a horizontal impulse P is imparted to the rod at a distance x=L / n from the mid-point of the rod (see figure), then the rod and string revolve together around the point O , with the ro

Step-by-step solution

Linear impulse F d t= momentum aligned & = m ( V _ cm -0 ) & P = m ( r _ cm ) & = m ( L + L 2 ) aligned P = m ( 3 ~L 2 ) ...(i) Angular impulse dt = angular momentum aligned & r Fdt = L & aligned & r Fdt = I ( -0), and I is moment of inertia about axis of rotation. & aligned & ( L + L 2 + x ) P = ( I _ cm + md ^2 ) &= ( mL ^2 12 + m ( L + L 2 )^2 ) aligned & ( 3 ~L 2 + x ) P = mL ^2 ( 1 12 + ( 3 2 )^2 ) aligned aligned ( 3 L 2 +x ) P= mL ^2 ( 7 3 ) ...(ii) Divide eq.-(i) & (ii) aligned & ( 3 L 2 +x )= L ( 7 3 ) ( 3