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Question

Physics Question on laws of motion

Two masses MM and mm are attached to a vertical axis by weightless thread of combined length l.l. They are set in rotational motion in a horizontal plane about this axis with constant angular velocity ω\omega . If the tensions in the thread are same during motion, the distance of M from the axis is,

A

MlM+m\frac{Ml}{M+m}

B

mlM+m\frac{ml}{M+m}

C

M+mMl\frac{M+m}{M}l

D

M+mml\frac{M+m}{m}l

Answer

mlM+m\frac{ml}{M+m}

Explanation

Solution

mrω2=MRω2m r \omega^{2}=M R \omega^{2} Put r=lRr=l-R m(lR)=MR\therefore m(l-R)=M R R=mlM+m\Rightarrow R=\frac{m l}{M +m}