Question

If, as soon as the angular speed is ω = 32 rad/s, the force is immediately removed and a mass m = 1.3 kg is dropped at r = 0.25 m on the disc and sticks there, what would be the new rotational velocity of the disc in rad/s?

Answer 1

Answer

given,

angular speed of disk = 32 rad/s

mass dropped, m = 1.3 Kg

radius, r = 0.25 m

new rotational velocity = ?

now,

Initial rotational inertia of the disk

$I_1 = \dfrac{1}{2}MR^2$

Assuming the mass and the radius of the disk is equal to 7.5 Kg and 0.85 m respectively

now,

$I_1 = \dfrac{1}{2}\times 7.5\times 0.85^2$

$I_1 = 2.71\ Kgm^2$

Rotational inertia after mass is dropped on it

$I_2 = \dfrac{1}{2}MR^2 + mr^2$

$I_2 = 2.71 + 1.3\times 0.25^2$

$I_2 = 2.79\ Kg.m^2$

using angular momentum conservation

 I₁ω₁ =  I₂ω₂

 2.71 x 32 = 2.79 x ω₂

 ω₂ = 31 rad/s

new angular velocity of the disk is   ω₂ = 31 rad/s