A uniform rod AB of mass 2 kg and length l = 100 cm is placed on a sharp support O such that AO = a = 40 cm and OB = b = 60 cm. A spring of force constant k = 600 N/m is attached to end B as shown in figure. To keep the rod horizontal, its end A is tied with a thread such that the spring is elongated by y = 1 cm. The reaction of support on the rod immediately after the thread is burnt is 5a N. Then a = ...
Physics · Mixed Chapter
JEE Mains

Question

A uniform rod AB of mass 2 kg and length l = 100 cm is placed on a sharp support O such that AO = a = 40 cm and OB = b = 60 cm. A spring of force constant k = 600 N/m is attached to end B as shown in figure. To keep the rod horizontal, its end A is tied with a thread such that the spring is elongated by y = 1 cm. The reaction of support on the rod immediately after the thread is burnt is 5a N. Then a = ...

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Immediately after the left thread is burnt, the forces other than support are weight mg=20 N and stretched spring force ky=600(0.01)=6 N, both down. The centre of mass is 0.10 m to the right of support O. Step 2: Moment of inertia about support: Iₒ=(1/12)ml² + m(0.1)²=(2/12)(1)²+2(0.01)=0.18667 kg·m². Step 3: Clockwise torque about O is (20)(0.1)+(6)(0.6)=5.6 N·m, so angular acceleration α=5.6/0.18667=30 rad/s². Initial CM vertical acceleration is 0.10α=3 m/s² down. Step 4: Vertical for…