Question

A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A mass of 1.1 kg is then dropped onto the edge of the wheel, where it sticks. What is the new rotational rate of the wheel

Answer 1

`\omega_f = 585.37 \ rev/s`

Step-by-step explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)= 800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

`I \omega_i = I' \omega_f`

`(Mr^2) * 800 = ( M r^2 + m r^2) \omega_f`

`M* 800 = ( M + m )\omega_f`

`3* 800 = (3+1.1)* \omega_f`

`2400 = (4.1)* \omega_f`

`\omega_f = 585.37 \ rev/s`