Calculating Angular Acceleration and Torque in a Rotating Motor System

[helpme101]

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
.000000002 m = radius
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]

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.

 

[helpme101]

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]

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 $x = x_0 + v_0 \, t + {\textstyle {1\over 2}}\alpha\, t^2$.

 

[helpme101]

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]

(Assume the motor pauses very briefly between steps and therefore starts each step from rest)

I don't understand this part of the problem statement. The motor applies a torque in each step, producing an acceleration. The angular velocity is now nonzero. When the motor rests, it stops applying a torque. Why does that mean the rotor comes to rest? There is no therefore about it. Presumably there is some resistance that the motor has to ovecome, and this will bring it to rest, but it will not come instantly to rest, so we don't know how long the rotation continues or what rate it got up to.