Mood:
crushed outIm thinking at least more than half of us wanna switch class/recitation or lab sections for Physics Two, so please post your requests here and hopfully someone else's needs will match what you have/want
| « | October 2007 | » | ||||
 |
||||||
| S | M | T | W | T | F | S |
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 7 | 8 | 9 | 10 | 11 | 12 | 13 |
| 14 | 15 | 16 | 17 | 18 | 19 | 20 |
| 21 | 22 | 23 | 24 | 25 | 26 | 27 |
| 28 | 29 | 30 | 31 | |||
crushed outIm thinking at least more than half of us wanna switch class/recitation or lab sections for Physics Two, so please post your requests here and hopfully someone else's needs will match what you have/want
for the first part,
rotational kinetic energy=.5*moment of inertia*(omega^2)
moment of inertia = .5mr^2 because u r dealing with a solid cylinder.
therefore, KE=.5*(.5mr^2)*(omega^2)
Find the tangental acceleration of circle A, because that will be the acceleration of the belt. tan acceleration = r*angular acceleration
then take this number, and find the angular acceleration of circle C by plugging in to the exact same equation, but use the radius of C and solve for alpha.
Once u have alpha for C, then the equation is just alpha=omega/t, and solve for t
For the "Torque on a merry go round," for the first part: the total torque(t)=the sum of all of the moments of inertia * the angular acceleration.
Find the acceleration by dividing your angular velocity ( remember to change to rad/s) by your time.
Then just add up the moments of inertia: for each child it's mr^2, but since the merry go round is a solid cylinder, it's moment of inertia is .5mr^2.
Hope that helps