The position vector of a particle of mass 2 kg is given as a function of time by ~r = (6 m) ˆı + (3 m/s)t ˆ . Determine the magnitude of the angular momentum of the particle with respect to the origin at time 3 s. Answer in units of kg m2 /s.

Explanation:

Given that mass of particle is 2kg

M=2kg

Given a position vector as a function of time

r=(6•i+3t•j)m

Angular momentum at t=3sec

Angular momentum is given as the cross product of position vector and momentum

I= r×p

Where r is position vector

And p is momentum

Momentum is given as Mass×Velocity

p=mv

The velocity can he determine from r by differentiating r with respect to t

r=(6•i+3t•j)m

v=r'

v=dr/dt= (0i+3j) m/s

v=3•j m/s

Therefore, p=mv=2×3j=6j

p=6•j kgm/s

Now, applying this to angular momentum

I=r×p

I= (6•i+3t•j) × 6•j

Note i×i=j×j=0, i×j=k, j×i=-k

I=6•i×6•j + 3t•j × 6•j

I=36•(i×j) +18t•(j×j)

I=36•k +18t•(0)

I=36•k kgm²/s

The angular momentum is

36•k kgm²/s

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards