কেন্দ্রমুখী ও কেন্দ্রবিমুখী বল

Two bodies A, B of masses m1,m2m_1, m_2 are knotted to a mass-less string at different points rotated along concentric circles in horizontal plane. The distances of A, B from common centre are 50cm, 1m. If the tensions in the string between centre to A and A to B are in the ratio 5:4, then the ratio of m1m_{1} to m2m_{2} is:

হানি নাটস

The tension will be more between center to A as it has to bear the centripetal force of 2 bodies.

applying force balance on the masses,

T1T2=m1ω2r1T_{1}-T_2=m_1 \omega^{2}r_1.........(1)

T2=m2ω2r2T_2=m_2 \omega^{2}r_2.........(1)

T2T1=45\Rightarrow \dfrac { { T }_{ 2 } }{ { T }_{ 1 } } =\dfrac { 4 }{ 5 } \quad

Now: r1=0.5{ r }_{ 1 } = 0.5, r2=1{ r }_{ 2 } =1

=> 54=m1(0.5)m2+1\dfrac { 5 }{ 4 } = \dfrac { { m }_{ 1 }(0.5) }{ { m }_{ 2 }+1 }

=> m1m2=2×14=12\dfrac{m_1}{m_2} = \dfrac{2\times 1}{4} = \dfrac{1}{2}

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