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A uniform disc rotating in a horizontal plane about vertical axis passing through its center is 20 rpm has a body mass 2kg kept at its center if the mass now slides to the edge what would the angular speed the disc be? mass of disc is 5 kg?
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A uniform disc rotating in a horizontal plane about vertical axis pass...
Problem:

A uniform disc rotating in a horizontal plane about vertical axis passing through its center is 20 rpm has a body mass 2kg kept at its center if the mass now slides to the edge what would the angular speed the disc be? mass of disc is 5 kg?

Solution:

Given,

Initial angular speed of the disc, w1 = 20 rpm

Mass of the disc, m1 = 5 kg

Mass at the center of the disc, m2 = 2 kg

Final angular speed of the disc, w2 = ?

Conservation of Angular Momentum:

When the mass at the center of the disc slides to the edge, there is no external torque acting on the system. Hence, the angular momentum of the system remains conserved.

L1 = L2

Where, L1 = Initial angular momentum of the system

L2 = Final angular momentum of the system

Calculation of Initial Angular Momentum:

Initial angular momentum of the system can be calculated as follows:

L1 = I1 w1

Where, I1 = Moment of inertia of the disc about its axis

I1 = (1/2) m1 r1^2

Where, r1 = Radius of the disc

r1 = (1/2) x Diameter = (1/2) x 0.2 m = 0.1 m

I1 = (1/2) x 5 kg x (0.1 m)^2 = 0.025 kg.m^2

L1 = 0.025 kg.m^2 x (20 rpm x 2π/60) = 0.52 kg.m^2/s

Calculation of Final Angular Momentum:

Final angular momentum of the system can be calculated as follows:

L2 = I2 w2

Where, I2 = Moment of inertia of the disc about its axis with mass at the edge

I2 = (1/2) m1 r2^2 + m2 d^2

Where, r2 = Radius of the disc with mass at the edge

r2 = Diameter = 0.2 m

d = Distance of the mass from the axis of rotation

d = r2 = 0.2 m

I2 = (1/2) x 5 kg x (0.2 m)^2 + 2 kg x (0.2 m)^2 = 0.36 kg.m^2

L2 = 0.36 kg.m^2 x w2

Equating L1 and L2, we get:

0.52 = 0.36 w2

w2 = 1.44 rad/s

Therefore, the final angular speed of the disc will be 1.44 rad/s.
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