A mass is whirled in a circular path with an angular momentum L. If the length of string and angular velocity, both are doubled, the new angular momentum is:

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Solution

The correct option is C 8L
: $L = m r^{2} ω$

$L \propto r^{2} ω$

$\frac{L_{1}}{L_{2}} = \frac{{r_{1}}^{2} ω_{1}}{{r_{2}}^{2} ω_{2}} = \frac{{r_{1}}^{2}}{4 {r_{1}}^{2}} . \frac{ω_{1}}{2 ω_{1}}$

$L_{2} = 8 L_{1}$

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A mass is whirled in a circular path with an angular momentum L. If the length of string and angular velocity, both are doubled, the new angular momentum is:

The correct option is C 8L: L=mr2ω L∝r2ω L1L2=r12ω1r22ω2=r124r12.ω12ω1 L2=8L1

The correct option is C 8L
: $L = m r^{2} ω$

$L \propto r^{2} ω$

$\frac{L_{1}}{L_{2}} = \frac{{r_{1}}^{2} ω_{1}}{{r_{2}}^{2} ω_{2}} = \frac{{r_{1}}^{2}}{4 {r_{1}}^{2}} . \frac{ω_{1}}{2 ω_{1}}$

$L_{2} = 8 L_{1}$