Question

Dur
A particle of mass M is moving in a straight line with uniform speed V parallel to X-axis in X-Y plane at a he
angular momentum about the origin is
/
(A) Zero
(B) mvh and is directed along positive Z-axis
(C) mvh and is directed along negative Z-axis
(D) mvh and is directed along positive X-axis
​

Answer 1

Velocity =v

i

^

position vector =a

i

^

+b

j

^

Angular momentum =m(a

i

^

+b

)

^

×v

i

^

and on doing cross product

=mvb

k

^

is the angular momentum of the particle

which represents in the direction of z-axis

In scalar form

=mvb

Answer 2

Answer:

The angular momentum is mvh and is directed along negative z axis

Explanation:

Given: A particle of mass M is moving in a straight line with uniform speed V parallel to X-axis

To find: The angular momentum of the origin

Solution:

Suppose the particle of mass 'm' is moving parallel to x-axis with constant velocity v and starting from B, where OB= b in the below figure.

At the time 't' let the particle be at p,

where x = OA = BP = vt.

y = b,

z = 0.

Components of velocity is given by

$V_{x} = v$ (It is parallel to x axis)

$V_{y} = 0$

$v_{z} = 0$

Equation should be written in vector form

L = r × p

    = r × mv   (where p = mv)

r and  v are the vectors so take m is a  constant

L = m ( r × v)

In matrix form

L = m × $\left[\begin{array}{ccc}r&j&k\\vt&b&o\\v&0&0\end{array}\right]$

L =  m k[vt × 0 - vb]

  = - mvbk

  = -mvhk

The angular momentum has magnitude m vb and is directed along negative z axis.

Therefore angular momentum remains constant.

Final answer:

In a plane the angular momentum has magnitude mvh and is directed along negative z axis

#SPJ2