# Angular Acceleration Help

## Homework Statement

The rotating subunits forms an approximately cylindrical axel with a diameter of 4 nm and total mass of 80,000 Daltons. The motor takes three steps per revolution with each step taking 0.067 s. We can model the rotating subunits as a rotating cylinder and use the properties described above to calculate:

A. The angular acceleration of the motor during each step (Assume the motor pauses very briefly between steps and therefore starts each step from rest)

B. The torque needed to generate the acceleration calculated in (a). Hint: you will need to determine the moment of inertia first.

## Homework Equations

A. Angular acceleration = change in angular velocity/ time interval
Angular velocity = 2 X pi radians X F
F=1/period

B. Torque = rF
Using the answer we found in part A the equation would be: angular acceleration = Torque/mr^2

## The Attempt at a Solution

A. Period = 0.067 s X 3 steps = .201 s
Frequency= 1/period = 1/.201s = 4.97512 s^-1
angular velocity = 2 X pi X 4.97512 s-1 = 31.2592 rad s-1
so angular acceleration = 31.2592 rad s-1 / 0.067 s - 0 s = 466.561 rads-2

B. 4nm=.000000004 m = diameter
80000 Daltons = 1.328 X 10 ^-22 kg

466.561 rad s^-2 = Torque/ (1.328 x 10^-22 kg) (.000000002 m)^2
= 2.47837 x 10^-37 Nxm

BvU
Homework Helper
Helllo SOS101 and welcome to PF !

You have no questions, but I do have a comment:

You treat both angular velocity and angular acceleration as constant. That is strange.

Helllo SOS101 and welcome to PF !

You have no questions, but I do have a comment:

You treat both angular velocity and angular acceleration as constant. That is strange.

I was hoping someone could check over my work. Sounds like I may have done something is wrong?

BvU
Homework Helper
Yes. If the angular acceleration is constant (and it seems the exercise is asking for a number), the angular velocity increases linearly and the angle goes (*) like ##{\textstyle {1\over 2}}\alpha\, t^2##. You have angle and t and can solve for ##\alpha##

(*) same as linear motion with constant acceleration formula ##x = x_0 + v_0 \, t + {\textstyle {1\over 2}}\alpha\, t^2##.

Yes. If the angular acceleration is constant (and it seems the exercise is asking for a number), the angular velocity increases linearly and the angle goes (*) like ##{\textstyle {1\over 2}}\alpha\, t^2##. You have angle and t and can solve for ##\alpha##

(*) same as linear motion with constant acceleration formula ##x = x_0 + v_0 \, t + {\textstyle {1\over 2}}\alpha\, t^2##.
I'm not sure if I am understanding. I should do (1/2)(.067s)^2 = alpha?

haruspex