Problem: A barbell consists of two small balls, each with mass m at the ends of a very low mass rod of length d. The barbell is mounted on the end of a low-mass rigid rod of length b. This apparatus is started in such a way that while the rod rotates clockwise with angular speed ω1, the barbell rotates clockwise about its center with an angular speed ω2. What is the total angular momentum of this system about point B?

Respuesta :

Answer:

mass of ball 1=m1

mass of ball 2=m2

velocity of ball=r1w1

velocity of ball 2=r2w2

Total angular momentum=m1*v1+m2*v2

but

v1=r1*w1

v2=r2*w2

Substitute values in above equation

Total angular momentum of the system=m1*r1*w1+m2*r2*w2

The total angular momentum of the system at point B  will be [tex]\rm L=m_1 r_1\omega_1 +m_2 r_2 \omega_2[/tex].

What is angular momentum?

The rotating counterpart of linear momentum is angular momentum also known as moment of momentum or rotational momentum.

The given data in the problem is;

m₁ is the mass of ball 1

m₂ mass of ball 2=m2

v₁ is the velocity of ball=r₁ω₁

v₂ is the velocity of ball 2=r₂ω₂

The total angular momentum of a system at point B  is given as;

[tex]\rm V_{total}= r_1\omega_1 + r_2 \omega_2 \\\\ \rm L=m_1 r_1\omega_1 +m_2 r_2 \omega_2[/tex]

Hence the total angular momentum of the system at point B  will be [tex]\rm L=m_1 r_1\omega_1 +m_2 r_2 \omega_2[/tex].

To learn more about the angular momentum refer to the link;

https://brainly.com/question/15104254