Two particles of mass 1 kg each are attached to a massless rod a distance of 1 m and 2m respectively from the axis of rotation. The angular speed of the system is 1 rad/s.

What is the angular momentum of the system? (in SI units)

a) 1

b) 2

c) 3

d) 4

Option (d)

The angular momentum of the system is 5 Kg-m²/s

Explanation:

Angular momentum: Angular momentum is the product of angular velocity and of moment of inertia.

Angular momentum is dented by L

And the unit of angular momentum is Kg-m²/s

L= I×ω  

I is momentum of inertia

ω is angular velocity.

The unit of angular velocity  is red/s.

Momentum of inertia : Momentum of inertia is the sum  square the perpendicular distance from the rotational axis to of product mass of each particle of the body.

The S.I unit of momentum of inertia is kg-m²

The angular momentum of a system is:

L= ∑I ω

  =∑ Mr²ω      [ Since ]

Given , M= 1 kg,  r₁=1 m , r₂= 2 m and   ω = 1 rad/s.

The angular momentum of the system is

L= Mr₁²ω+Mr₂²ω

  = Mω(r₁²+r₂²)

 = {1×1 (1²+2²)}

=5 Kg-m²/s

given data:

determine the horse power (h.p) = ?

force applied to move the box(f) = 10 pounds,

box lifted from a distance         (d) = 3 feet

time taken to lift                         (t) = 10 sec,

the horse power is defined as amount of work required to move a load. the horse power is equal to 550 foot-pound per second

                  1 h.p   = 550 foot-pound/second

it can also be defined as the the load that need to be move a body for certain distance in a certain amount of time.

mathematically,

                      h.p = (force × distance)/ time

                            = (10 × 3)/ 10

                            = 30/10

                            = 3 foot -pound/sec

hence, the horse power is needed to move the box is 3 foot-pound/second

important note: 1 h.p = 550 foot-pound/second = 746 watts

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