Download the app

Questions  

For a particle of a mass 100 gm, position and velocity at any instant are given as 10i^+6j^ cm and v = 5i^ cm/s. Calculate the angular momentum about the point (1, 1) cm.

a
25 × 10-5(-k^)  kgm2s-1
b
30× 10-5(-k^) kgm2s-1
c
3 × 10-3(-k^)kgm2s-1
d
25× 10-4(-k^) kgm2s-1

detailed solution

Correct option is A

The position vector of particle w.r.t. (1,1) is r→=10-1i^+6-1j^ = 9i^+5j^L→ = m(r→ × v→) = 1001000[9i^+5j^100]×5i^100= 25 ×10-5(-k^) kgm2s-1

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

A particle of mass m = 5 units is moving with a uniform speed v=32, units in the XOY plane along the line y = x + 4. The magnitude of the angular momentum of the Particle about the origin is


phone icon
whats app icon