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QJEEM19Ph6651: JEE Main 2019 Physics - Class 11 JEEM19Ph6651 Solution: Two masses m and are connected at the two ends of a massless…

Two masses m and m2{m \over 2} are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k, at the centre of mass of the rod-mass system(see figure). Because of torsional constant k, the restoring torque is τ\tau = k θ\theta for angular displacement θ\theta . If the rod is rotated by θ\theta 0 and released, the tension in it when it passes through its mean position will be :

JEE Main 2019 (Online) 9th January Morning Slot Physics - Center of Mass and Collision Question 100 English

Multiple Choice Options

  • A 3kθ02l{{3k{\theta _0}^2} \over l}
  • B 2kθ02l{{2k{\theta _0}^2} \over l}
  • C kθ02l{{k{\theta _0}^2} \over l}
  • D kθ022l{{k{\theta _0}^2} \over {2l}}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 9th January Morning Slot Physics - Center of Mass and Collision Question 100 English Explanation 1
∴\therefore     r1=(n2m+m2)l{r_1} = \left( {{{{n \over 2}} \over {m + {m \over 2}}}} \right)l

r1=l3{r_1} = {l \over 3}

When rod is rotated by θ\theta 0 and released, then it look this.

JEE Main 2019 (Online) 9th January Morning Slot Physics - Center of Mass and Collision Question 100 English Explanation 2
After roating θ\theta 0 and released, when the rod go through mean position it will have maximum speed.

Vmax == A ω\omega

== r1 θ\theta 0 ω\omega

== (l3θ0)ω\left( {{l \over 3}{\theta _0}} \right)\omega

We know,

ω\omega =KI= \sqrt {{K \over {\rm I}}}

and     I=m(l3)2+m2(2l3)2{\rm I} = m{\left( {{l \over 3}} \right)^2} + {m \over 2}{\left( {{{2l} \over 3}} \right)^2}

=ml29+4ml218= {{m{l^2}} \over 9} + {{4m{l^2}} \over {18}}

=2ml2+4ml218= {{2m{l^2} + 4m{l^2}} \over {18}}

=ml23= {{m{l^2}} \over 3}

∴\therefore     ω=3Kml2\omega = \sqrt {{{3K} \over {m{l^2}}}}

∴\therefore    Tension in the rod when it passes through the mean position is,

T = mvmax⁡2r1{{mv_{\max }^2} \over {{r_1}}}

=m(l3θ0ω)2l3= {{m{{\left( {{l \over 3}{\theta _0}\omega } \right)}^2}} \over {{l \over 3}}}

=m(l3)2θ02ω2ℓ3= {{m{{\left( {{l \over 3}} \right)}^2}\theta _0^2{\omega ^2}} \over {{\ell \over 3}}}

=m(l3)θ02(3Kml2)= m\left( {{l \over 3}} \right)\theta _0^2\left( {{{3K} \over {m{l^2}}}} \right)

=Kθ02l= {{K\theta _0^2} \over l}

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