Question

A disk with a rotational inertia of 8.0 kg * m2 and a radius of 1.6 m rotates on a frictionless fixed axis perpendicular to the disk faces and through its center. A force of 10.0 N is applied tangentially to the rim. The angular acceleration of the disk is:

Answer 1

Answer:

α = 2  rad/s²

Explanation:

Newton's second law for rotation:

τ = I * α   Formula  (1)

where:

τ : It is the torque applied to the body.  (N*m)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Data

I =  8.0 kg * m²   :moment of inertia of the disk

R =  1.6 m : radius of the disk

F = 10.0 N : tangential force applied to the disk

Torque applied to the disk

The torque is defined as follows:

τ = F*R

τ = 10.0 N* 1.6 m

τ = 16 N*m

Angular acceleration of the disk ( α  )

We replace data in the formula (1):

τ = I * α

16 = 8 *α

α = 16 / 8

α = 2  rad/s²