p=mvj^=2×vj^=2×0.6×12=7.2×2=14.4j^(v=rω)
r=(0.8k^+0.6i^)⇒L=r×p
=(0.6i^+0.8k^)×14.4j^
=0.6×14.4k^−0.8×14.4i^
=14.4(0.6k^−0.8i^)
⇒∣L∣=14.4(0.6)2+(0.8)2
=14.40.36+0.64
=14.4×1
=14.4kgm2s−1
A particle of mass 2kg is on a smooth horizontal table and moves in a circular path of radius 0.6m. The height of the table from the ground is 0.8m. If the angular speed of the particle is 12rads−1 , the magnitude of its angular momentum about a point on the ground right under the center of the circle is:
Held on 11 Apr 2015 · Verified 6 Jul 2026.
14.4kgm2s−1
11.52kgm2s−1
20.16kgm2s−1
8.64kgm2s−1
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