Question

A disk with a rotational inertia of 2.0 kg m2 and a radius of 1.6 m is free to rotate about a frictionless axis perpendicular to the disk's face and passing through its center. A force of 5.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk as a result of this applied force?

a) 4.0 rad/s2
b) 1.0 rad/s2
c) 2.0 rad/s2
d) 0.40 rad/s2
e) 0.80 rad/s2

Answer 1

a) 4.0 rad/s2

Step-by-step explanation:

• For rigid bodies, Newton's 2nd law becomes :

τ = I * α (1)

where τ is the net external torque applied, I is the rotational inertia

of the body with respect to the axis of rotation, and α is the angular

acceleration caused by the torque.

• At the same time, we can apply the definition of torque to the left side of (1), as follows:

`\tau = F*r*sin \theta (2)`

where τ = external net torque applied by Fnet, r is the distance

between the axis of rotation and the line of Fnet, and θ is the

angle between both vectors.

In this particular case, as Fnet is applied tangentially to the disk, Fnet

and r are perpendicular each other.

• Since left sides of (1) and (2) are equal each other, right sides are equal too, so we can solve for the angular acceleration as follows:

`\alpha = (F*r)/(I) = (5.0N*1.6m)/(2.0 kg*m2) = 4.0 rad/s2 (3)`