A uniform rod of mass m is hinged (ignore friction at hinge) as shown and is released from a horizontal position. The angular velocity of the rod as it passes the vertical position is given by √(xg/ℓ). Find the value of x.
Physics · Mixed Chapter
JEE Mains

Question

A uniform rod of mass m is hinged (ignore friction at hinge) as shown and is released from a horizontal position. The angular velocity of the rod as it passes the vertical position is given by √(xg/ℓ). Find the value of x.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The hinge is located ℓ/3 from the left end. The rod centre of mass is at ℓ/2, so its distance from hinge is d=ℓ/2−ℓ/3=ℓ/6. Step 2: Moment of inertia about hinge is I=(mℓ²/12)+m(ℓ/6)²=mℓ²/9. Step 3: When the rod reaches the vertical downward position, its centre of mass has fallen by ℓ/6. Energy conservation gives ½Iω²=mg(ℓ/6). Step 4: ½(mℓ²/9)ω²=mgℓ/6, so ω²=3g/ℓ and ω=√(3g/ℓ). Hence x=3.