Recent content by la673

how would u work out the final kinetic energy here? i tried using k_e=.5*I*(omega)^2 but I is an unknown, as to find I, u use I=kmr2, so I=(1/2)*m*(.03)^2 (as this is a solid cylinder) and i have no idea how to work out m. wait can u use f=ma? so 40 n = m* a, where a =r*alpha, or=2.499...

heh i got it, but i knew that, i just didnt realize i needed to use torque= I*alpha, nstead of momentum/time. then u solve alpha down into terms of intital and final angular velocity over time, and bingo

nd so by using total time is 12.9s, u use the formula omega_max=omega_intital+alphja*t to find the angular deccelration? so 107.2=29+alpha_deceleration*12.9, so alpha is around 6.06? please tell me if this look right, or am i doing it all wrong? -edit: lol srry didnt even notice the edit...

nd so the chnage in time is 10.6 s after the intital 2.3 s?

wiat, os 14.78= 156.63/change in time?

change in angle/ change in time, but i can't figure out after what time it stops.

107.2, usin omega=omega_0+alpha_o*t

heh i got the first 2 easy, but I am totally lost on the third, how do u work out the angular momentum?

At time t=0 a grinding wheel has an angular velocity of 29.0 rad/s. It has a constant angular acceleration of 34.0 rad/s^2 until a circuit breaker trips at time t = 2.30 s. From then on, the wheel turns through an angle of 434 rad as it coasts to a stop at constant angular deceleration. now...

yeh, knowler673@hotmail.com

i kno wat u mean, now I am stuck on the next question... got 12.61 m/s which is wrong... then i used 13 (cause it says to nearest integer) nd its also wrong

yup, so do u i take it? nd thanks for that it was driving me crazy... couldn't work out where i was wrong

please help if u can. Consider the following two-car accident: Two cars of equal mass collide at an intersection. Driver E was traveling eastward, and driver N, northward. After the collision, the two cars remain joined together and slide, with locked wheels, before coming to rest. Police on...