At one instant, force F=4.0j^Nacts on a 0.25kgobject that has position vector r=(2.0i^-2.0k^)mand velocity vectorv=(-5.0i^+5.0k^)m/sAbout the origin and in unit-vector notation, (a) What is the object’s angular momentum? (b) What is the torque acting on the object?

Short Answer

Expert verified
  1. The angular momentum of the object l = 0 kg.m2/s
  2. The torque acting on the object is τ=8.0N.mi^+8.0N.mj^

Step by step solution

01

Identification of given data

The mass of the object is m = 0.25 kg

The force acting on the object isF=4.0j^N

The position vector isr=2.0mi^-2.0mk^

The velocity vector isv=-5.0m/si^+5.0m/sk^

02

To understand the concept angular momentum

The problem deals with the calculation of angular momentum. The angular momentum of a rigid object is product of the moment of inertia and the angular velocity. It is analogous to linear momentum. It also deals with the torque. It is a measure of the force that can cause an object to rotate about an axis

Formulae:

l=mr×vτ=r×F

03

(a) Determining the angular momentum of the object

Let position vector is r=xi^+yj^+zk^and velocity vector is v=vxi^+vyj^+vzk^.

The cross product of the position vector and velocity vector is,

r×v=yvz-zvyi^+zvx-xvzj^+xvy-yvxk^

In the given position and velocity vector, y = 0 and vy = 0m/s. Then,

r×v=zvx-xvzj^

The angular momentum of the object with position vector and velocity vector is,

l=mr×v=mzvx-xvzj^=0.25kg-2.0m-5.0m/s-2.0m5.0m/sj^=0kgm2s

04

(b) Determining the torque acting on the object

To find torque acting on the object, force acting on it will be F=Fxi^+Fyj^+Fzk^.

The expression of torque is,

τ=r×F=yFz-zFyi^+zFx-xFzj^+xFy-yFxk^

In the given position and force vector, y= 0m and

Fx=Fz=0Nτ=-zFyi^+xFyk^=--2.0m4.0Ni^+2.0m4.0Nk^=8.0N.mi^+8.0N.mk^

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