Two particles of mass m each are fixed to a massless rod of length 2l, hinged smoothly at one end to a ceiling. For small oscillations in a vertical plane, the equivalent simple-pendulum length is $\frac{xl}{3}$ .Find x.
Your answer

Question

Two particles of mass m each are fixed to a massless rod of length 2l, hinged smoothly at one end to a ceiling. For small oscillations in a vertical plane, the equivalent simple-pendulum length is $\frac{xl}{3}$ .Find x.

Your answer

G1 Physics · Accepted Answer

The two masses m are at distances l and 2l from the hinge. Thus I=m l²+m(2l)²=5ml². Their combined centre of mass is h=(l+2l)/2=3l/2 from the hinge and total mass is 2m. Equivalent simple-pendulum length Lₑ=I/(2m h)=5ml²/[2m(3l/2)]=5l/3. Since it is written as xl/3, x=5.00.

Two particles of mass m each are fixed to a massless rod of length 2l, hinged smoothly at one end to a ceiling. For small oscillations in a vertical plane, the equivalent simple-pendulum length is x l 3 \frac{xl}{3} 3 x l ​ .Find x.

Two particles of mass m each are fixed to a massless rod of length 2l, hinged smoothly at one end to a ceiling. For small oscillations in a vertical plane, the equivalent simple-pendulum length is x l 3 \frac{xl}{3} 3 x l .Find x.