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Standard XII

Physics

Deducing Mathematically

A rod of mass...

Question

A rod of mass $m$ and length $2 R$ is fixed along the diameter of a ring of same mass $m$ and radius $R$ as shown in figure.The combined body is rolling without slipping along x-axis. Find the angular momentum about z-axis.

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Solution

Centre of mass of both lies at the centre of ring
$I_{c} = m R^{2} + \frac{m (2 R)^{2}}{12}$
$= \frac{4}{3} m R^{2}$
$L = (2 m v r_{⊥} + I_{c} ω$
$= (2 m v R) + \frac{4}{3} m R^{2} (\frac{v}{R})$
$\frac{10}{3} m v R$
TIlls is clockwise, or along negative z-axis
$∴ \to L = (- \frac{10}{3} m v R)^k$

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A rod of mass m and length 2R is fixed along the diameter of a ring of same mass m and radius R as shown in figure.The combined body is rolling without slipping along x-axis. Find the angular momentum about z-axis.

Centre of mass of both lies at the centre of ringIc=mR2+m(2R)212=43mR2L=(2mvr⊥+Icω=(2mvR)+43mR2(vR)103mvRTIlls is clockwise, or along negative z-axis∴→L=(−103mvR)^k

A rod of mass $m$ and length $2 R$ is fixed along the diameter of a ring of same mass $m$ and radius $R$ as shown in figure.The combined body is rolling without slipping along x-axis. Find the angular momentum about z-axis.