A thin uniform rod ''AB'' of mass m and length L is hinged at one end A to the level floor. Initially it stands vertically and is allowed to fall freely to the floor in the vertical plane. The angular velocity the rod, when its end B strikes the floor is (g is acceleration to gravity)

Question

A thin uniform rod ''AB'' of mass m and length L is hinged at one end A to the level floor. Initially it stands vertically and is allowed to fall freely to the floor in the vertical plane. The angular velocity the rod, when its end B strikes the floor is (g is acceleration to gravity) (A) mg/L (B) (mg/3L)(1/2) (C) g/L (D) (3g/L)(1/2)

Tardigrade Admin · Accepted Answer

The velocity with which the point 'B' strikes the ground is V = √3rg but ν = r ω ∴ r2 ω2 = 3rg ⇒ ω = √(3g/r) = √(3g/L)

Q. A thin uniform rod ''AB'' of mass m and length L is hinged at one end A to the level floor. Initially it stands vertically and is allowed to fall freely to the floor in the vertical plane. The angular velocity the rod, when its end B strikes the floor is (g is acceleration to gravity)

mg/L

$(m g /3 L)^{\frac{1}{2}}$

g/L

$(3 g / L)^{\frac{1}{2}}$

The velocity with which the point 'B' strikes the ground is $V = 3 r g$
but $ν = r ω$
$∴$ $r^{2} ω^{2} = 3 r g \Rightarrow ω = \frac{3 g}{r} = \frac{3 g}{L}$

Q. A thin uniform rod ''AB'' of mass m and length L is hinged at one end A to the level floor. Initially it stands vertically and is allowed to fall freely to the floor in the vertical plane. The angular velocity the rod, when its end B strikes the floor is (g is acceleration to gravity)

mg/L

(mg/3L)21​

g/L

(3g/L)21​

The velocity with which the point 'B' strikes the ground is V=3rg​ but ν=rω ∴ r2ω2=3rg⇒ω=r3g​​=L3g​​

$(m g /3 L)^{\frac{1}{2}}$

$(3 g / L)^{\frac{1}{2}}$

The velocity with which the point 'B' strikes the ground is $V = 3 r g$
but $ν = r ω$
$∴$ $r^{2} ω^{2} = 3 r g \Rightarrow ω = \frac{3 g}{r} = \frac{3 g}{L}$