Topic: HW 9a #2
A fish takes the bait and pulls on the line with a force of 1.5 N. The fishing reel, which rotates
without friction, is a cylinder of radius 0.045 m and mass 0.62 kg.
(a) What is the angular acceleration of the fishing reel?
Student A:
First I found Torque=F*r
Second I found Inertia=m*r^2
Then I put both answers into alpha=T/I
I got the wrong answer, can anyone put me in the right direction. I thought for sure this is what
I needed to do. Thanks!
Dr. Man: Right, a cylinder is like a disc, refer to your book or my lecture notes to find
its rotational inertia
I= 1/2 M R^2
Basically, force causes acceleration, a=F/m
and torque causes angular acceleration:
alpha = torque / moment inertia
Once you find alpha, since it is constant, (because the torque is constant, you can find the
angle you covered (delta_theta) during time t.
You know alpha, you know t, you know it started at zero angular acceleration. It’s
straightforward to find total angle coved. (delta_theta)
You can then relate the angle it covered with the arc length it covered
by multiplying ______?____
BTW, alpha = torque / moment inertia
This equation was worked through in the sample and demo I gave during the class:
Finding the angular acceleration of your arm under torque caused by Mg, and finding the
acceleration at your fingertip, (when you let it go and drop your arm, your fingertip dropped
faster than the pencil in your hand... )
Topic: HW 9a #4
3 kg mass and a 4 kg mass are attached to either end of a 3 m long massless rod.
a.) Find the center of mass of the system.
m, from the 3 kg mass.
Find the rotational inertia (I) of the system when rotated about:
b.) the end with the 3 kg mass.
c.) the end with the 4 kg mass.
d.) the center of the rod.
e.) the center of mass of the system.
(Compare this to parts b-d. Is this what you expect?)
f.) If the system is rotated about the center of mass by a force of 10 N acting on the 4 kg mass,
(perpendicular to the rod), what will the size of the angular acceleration of the system be?
rad/s2
Dr. Man:
Some asked “didn’t we find the center of mass of the system in part a? and I’m not sure how to
find the center of rod."
The rod has no mass. the center of mass of the system (m1 and m2) can be easily found in part a
Question d asks you to find the moment of inertia, I, using the center of the rod to be the
rotation center. The center of rod is just half length of the rod.
I = sum of (mr^2) for all masses. When you choose different rotation center, r for those
masses are all different.
Question e asks you to find the moment of inertia using center of mass of the system to be
the rotation center.
The center of mass of the system is what you find previously in part a.
Question f asks you to use the rotational inertia you found in part e to find the alpha.
Student A:
If the system is rotated about the center of mass by a force of 6 N acting on the 4 kg mass,
(perpendicular to the rod), what will the size of the angular acceleration of the system be?
I got from 4d that my inertia will be 11.25 and the rod is 3m long meaning that the radius should
be 1.5. I multiplied that by the force and divided by 11.25 but my answer is incorrect. Am I
not getting the right radius to use? I tried using the COM for the whole system and that didn't
seem to work so i am stumped. Any help on where i am missing would be helpful.
Dr. Man: Read related posts here first. :)
In another post, I already said:
Question e asks you to find the moment of inertia using center of mass of the system
to be the rotation center.
The center of mass of the system is what you find previously in part a.
Question f asks you to use the rotational inertia you found in part e to find the alpha