Question

Moment of inertia of a hollow thin walled sphere

Answer 1

A hollow, thin-walled sphere of mass 13.0kg and diameter 47.0cm is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by\[\theta(t)=At^2+Bt^4\]where A has numerical value 1.10 and B has numerical value 1.60. At the time 4.00s , find the angular momentum of the sphere. At the time 4.00s , find the net torque on the sphere.

Answer 2

so for the first part i found that the moment of inertia for this object would be \[I=\frac{4}{3}MR^2=0.957\]then i just plugged 4s into to equation given\[\theta(4)=1.10(4)^2+1.60(4)^4=427.2\]then i used some kinematics\[427.2=\frac{1}{2}\alpha(4)^2\]\[\alpha=53.4\]then used another equation\[\omega=\omega_0+\alpha t \rightarrow \omega=(53.4)(4)=213.6\]and then once again i used another equation\[L=I\omega=(0.957)(213.6)=204.4\]now i'm trying to find the second part and i know that \[\tau_{net}=I_{net}\alpha\]so i just plugged in\[\tau_{net}=(0.957)(53.4)=51.1\]what did i do wrong?

Answer 3

Are you sure the I = 4/3MR^2 for a thin walled sphere?

Answer 4

I have derived it from first principles and get a different formula

Answer 5

i just assumed that it would be 2/3mr^2+2/3mr^2

Answer 6

No, it is just I = 2/3 MR^2

Answer 7

i used the parallel axis theorem since it's a thin walled hollow sphere.

Answer 8

hang on... http://www.engineeringtoolbox.com/moment-inertia-torque-d_913.html

Answer 9

\[I=I_{com}+Md^2\]

Answer 10

look under thin walled hollow sphere

Answer 11

hmmm... oh... i guess i used something way off. i'm surprised the first answer i got was so close to the correct answer. i got 204 and the answer ended up being 200. i'll redo this when i get back to my apartment. right now I'm on the bus and it's cramped in here.

Answer 12

lol...ok

Answer 13

Something else to consider, you assumed alpha is constant...it is not.
Let me know if you can see why it is not constant

Answer 14

Hint: theta is a function of time (given in problem)
omega (by defintion is ) = dtheta/dt
alpha (by definition is = domega/dt

Answer 15

it's not constant because the delta theta will grow quickly since there is a t^2 and a t^4 in the equation

Answer 16

yes...something very important to keep in mind the following definitions always apply:
omega = dtheta/dt
alpha = domega/dt

BUT the kinematic equations such as omega = omega(initial) + alpha*t
ONLY apply when alpha is a constant

Make sense?

Answer 17

yea

Answer 18

Super...they love to put questions like this on engineering exams!

Answer 19

yea. luckily i have it on homework. i would have done this on the exam.

Answer 20

i took the derivative of theta(t) and plugged in the given time and then multiplied it by 2/3M(d/2)^2 and got my L then i took the second derivative of theta(t) and did the same thing as the first one to find my net torque. i looked back at my notes and saw what you told me. thanks for helping me.

Answer 21

your welcome!