A particle has the position vector r-2-k- and the linear momentum p-2-k- Then- its angular momentum about the origin is

Angular momentum, L=r×p =î amp; ĵ amp; k̂ nbsp; 1 amp; 2 amp; 1 nbsp; 2 amp; 1 amp;1 nbsp; =î( 2+1) ĵ(1 2)+k̂( 1+4) = î+ĵ+3k̂ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Cross Product

A particle ha...

Question

A particle has the position vector $\to r =^i - 2^j +^k$ and the linear momentum $\to p = 2^i -^j +^k$ . Then, its angular momentum about the origin is

-^i +^j - 3^k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-^i +^j + 3^k

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

^i +^j + 3^k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^i -^j - 5^k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $-^i +^j + 3^k$
Angular momentum, $\to L = \to r \times \to p$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & - 2 & 1 2 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$=^i (- 2 + 1) -^j (1 - 2) +^k (- 1 + 4)$
$= -^i +^j + 3^k$

Suggest Corrections

Similar questions

Q. A particle has the position vector

\to r =^i - 2^j +^k

and the linear momentum

\to p = 2^i -^j +^k

. Then, its angular momentum about the origin is

Q. The position of particle is given by

\to r =^i + 2^j -^k

and it's linear momentum is given by

\to p = 3^i + 4^j - 2^k

. Then it's angular momentum about the origin is perpendicular to

Q. The position of a particle is given by

\to r = (^i + 2^j -^k) m

and linear momentum as

\to p = (3^i + 4^j - 2^k) kg-m/s

. The angular momentum of the particle about the origin

(0, 0, 0)

is perpendicular to

The Vector $- - \to O A = - 2^i +^j + 2^k$ is rotated through an angle $90^{\circ}$ about origin (O)In new Position it passes through $^i +^k$ . Then the vector in new position is

Q. A particle of mass

2 k g

located at the position

(^i +^j) m

has a velocity

2 (^i -^j +^k)

m / s

. Its angular momentum about z-axis in

k g m^{2} / s

is :

Join BYJU'S Learning Program

Explore more

Cross Product

Standard XII Physics

Join BYJU'S Learning Program

A particle has the position vector →r=^i−2^j+^k and the linear momentum →p=2^i−^j+^k. Then, its angular momentum about the origin is

The correct option is B −^i+^j+3^kAngular momentum, →L=→r×→p =∣∣ ∣ ∣∣^i^j^k1−212−11∣∣ ∣ ∣∣ =^i(−2+1)−^j(1−2)+^k(−1+4) =−^i+^j+3^k

A particle has the position vector $\to r =^i - 2^j +^k$ and the linear momentum $\to p = 2^i -^j +^k$ . Then, its angular momentum about the origin is

The correct option is B $-^i +^j + 3^k$
Angular momentum, $\to L = \to r \times \to p$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & - 2 & 1 2 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$=^i (- 2 + 1) -^j (1 - 2) +^k (- 1 + 4)$
$= -^i +^j + 3^k$