Question

A disk of radius 1.6 m and mass 3.8 kg rotates about an axis through its center. A force of 18.4 N is applied tangentially to the edge of the hoop, causing it to rotate counterclockwise. Assuming the disk is initially at rest, after 3 s... a) what is the radial acceleration of a point halfway between the axis and the edge of the disk

Answer 1

263.8 m/s2

Step-by-step explanation:

Assume this is a solid disk, we can find its moments of inertia:

`I = mr^2/2 = 3.8*1.6^2/2 = 4.864 kgm^2`

The torque T generated by force F = 18.4N is:

`T = Fr = 18.4*1.6 = 29.44 Nm`

So the angular acceleration of the disk according to Newton's 2nd law is:

`\alpha = T/I = 29.44 / 18.4 = 6.05 rad/s^2`

If the disk starts from rest, then after 3s its angular speed is

`\omega = \alpha \Delta t = 6.05*3 = 18.16 rad/s`

And so its radial acceleration at this time and half way from the center to the edge is:

`a_r = \omega^2(r/2) = 18.16^2*(1.6/2) = 263.8 m/s^2`

Note that this value is the same anywhere

Answer 2

At the edge, angular acceleration = 3.025 rad/s2

At halfway = 13.11 rad/s2

Step-by-step explanation:

Detailed explanation and calculation is shown in the image below