Question

A cylinder with moment of inertia 41.8 kg*m^2 rotates with angular velocity 2.27 rad/s on a frictionless vertical axle. A second cylinder, with moment of inertia 38.0 kg*m^2, initially not rotating, drops onto the first cylinder and remains in contact. Since the surfaces are rough, the two eventually reach the same angular velocity. Calculate the final angular velocity.

Answer 1

The resultant angular velocity = 1.19 rad/s

Explanation:

When any body rotates about its axis , the angular momentum of the body remains constant .

Thus the product of moment  of inertia and its angular velocity remains constant .

In first case

The moment of inertia of cylinder = 41.8 kg m²

and Angular velocity = 2.27 rad/s

Thus angular momentum L₁ = 41.8 x 2.27 N-m s

In the second case

The moment of inertia = 41.8 + 38.0 = 79.8 kg m²

Suppose the angular velocity = ω

Thus angular momentum L₂ = 79.8 x ω

But according to principle of conservation of momentum

L₁ = L₂

41.8 x 2.27 = 79.8 x ω

Thus ω = $\frac{41.8x2.27}{79.8}$

ω = 1.19 rad/s

Answer 2

Answer:

Final Angular Velocity=$\omega$=1.189 rad/s

Explanation:

Given Data:

Moment of Inertia of 1st cyclinder=$I_1=41.8 kg/m^{2}$

Angular Velocity of 1st cyclinder=$\omega_1$=2.27 rad/s

Moment of Inertia of After contact=$I_2=(41.8+38) kg/m^{2}=79.8 kg/m^{2}$

Required:

Final Angular velocity =$\omega$=?

Formula:

Angula Momentum=L=$I\omega$

Solution:

According to the conservation of angular momentum:

$L_1=L_2$

$I_1 \omega_1=I_2\omega\\41.8*2.27=(41.8+38)*\omega\\\omega=\frac{41.8*2.27}{79.8}\\\omega= 1.189\ rad/s$

Final Angular Velocity=$\omega$=1.189 rad/s