Question

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 , the barbell rotates clockwise about its center with an angular speed wg. What is the total angular momentum of this system about point B?

Answer 1

The total angular momentum of the system about point B is $L=m_1r_1\omega_1+m_2r_2\omega_2$

Angular momentum, also known as moment of momentum or rotational momentum, is the rotating counterpart of linear momentum.

A rigid object's angular momentum is defined as the product of its moment of inertia and its angular velocity. If there is no external torque on the object, it is analogous to linear momentum and is subject to the fundamental constraints of the conservation of angular momentum principle. The vector quantity angular momentum It is derived from the expression for a particle's angular momentum.

Given,

mass of ball 1 = m1

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 is given as;

$V_{total}=r_1\omega_1+r_2\omega_2\\\\L=m_1r_1\omega_1+m_2r_2\omega_2$

Hence the total angular momentum  will be $L=m_1r_1\omega_1+m_2r_2\omega_2$

To learn more about angular momentum refer here

brainly.com/question/29512279

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