Question

A light-weight potter’s wheel, having a moment of inertia of 24 kg m2 is spinning freely at 40.0 rpm. The potter drops a small but dense lump of clay onto the wheel, where it sticks a distance 1.2 m from the rotational axis. If the subsequent angular speed of the wheel and clay is 32 rpm, what is the mass of the clay?

Answer 1

m=4.03 kg

Step-by-step explanation:

Given that

r= 1.2 m

I=24 kg.m²

N₁ = 40 rpm

`\omega_1=(2\pi N_1)/(60)\ rad/s`

`\omega_1=(2\pi * 40)/(60)\ rad/s`

ω₁= 4.1 rad/s

N₂=32 rpm

`\omega_2=(2\pi * 32)/(60)\ rad/s`

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