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

Answer:

$\omega_f = 585.37 \ rev/s$

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

$I \omega_i = I' \omega_f$

$(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f$

$M\times 800 = ( M + m )\omega_f$

$3\times 800 = (3+1.1)\times \omega_f$

$2400 = (4.1)\times \omega_f$

$\omega_f = 585.37 \ rev/s$