Answer :
Answer:
m=4.03 kg
Explanation:
Given that
r= 1.2 m
I=24 kg.m²
N₁ = 40 rpm
[tex]\omega_1=\dfrac{2\pi N_1}{60}\ rad/s[/tex]
[tex]\omega_1=\dfrac{2\pi \times 40}{60}\ rad/s[/tex]
ω₁= 4.1 rad/s
N₂=32 rpm
[tex]\omega_2=\dfrac{2\pi \times 32}{60}\ rad/s[/tex]
ω₂ = 3.3 rad/s
Lets take mass of clay = m kg
Initial angular momentum L₁
L₁ = Iω₁
Final linear momentum L₂
L₂= I₂ ω₂
I₂ = I + m r²
The is no any external torque that is why angular momentum will be conserve
L₁ = L₂
Iω₁ = I₂ ω₂
Iω₁ = ( I + m r²) ω₂
Now by putting the all values
Iω₁ = ( I + m r²) ω₂
24 x 4.1 = (24 + m x 1.2²) x 3.3
24 x 4.1 = (24 + 1.44 m) x 3.3
29.81 = 24 + 1.44 m
1.44 m = 5.81
m=4.03 kg