Question

A uniform solid disk with a mass of 24.3 kg and a radius of 0.314 m is free to rotate about a frictionless axle. Forces of 90 and 125 N are applied to the disk in the same horizontal direction but one is applied to the top and the other is applied to the bottom. What is the angular acceleration of the disk (in rad/s2)?

Answer 1

Answer:

α = 9.18 rad/s²

Explanation:

given,

mass of the solid disk = 24.3 Kg

radius of the disk = 0.314 m

Force, F₁ = 90 N

           F₂ = 125 N

net force acting on the disk

F = 125 - 90

F = 35 N

Torque

τ = F . r

τ = 35 x 0.314

τ = 11 N.m

we know that

τ = I α

moment of inertia of the solid disk

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

$I = \dfrac{1}{2}\times 24.3\times 0.314^2$

   I = 1.198 kg.m²

now,

11 = 1.198 x α

α = 9.18 rad/s²

the angular acceleration of the disk is equal to 9.18 rad/s²