Recent content by saturn67

ok thank a lot alphysicist that is a big help XD yep i got it w= 4.8 rad/sec

humm soo L= R cross P p= mv so L=m1*v1*r is right?

how to relate I1*w1i to bullet?? i know I2w2i= 0 since stick is at rest

angular momentum conserved? I*wi = I*wf but bullet don't have angular momentum? so 0=I*wf so it don't make sense >.<

Homework Statement A uniform stick 1.0 m long with a total mass of 220 g is pivoted at its center. A 3.2 g bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250 m/s and leaves at 140 m/s With what angular speed is the stick spinning after the...

WOOT it work BIG thank! Doc Al for help me

T1 - M1*g = + M1*a (found T1) T2 - M2*g = - M2*a (found T2) I=0.000864 T1*R + T2*R + frictional torque = I*alpha frictional =0.5mN

Vf=Vo+at a=0.034328m/s^2 right?

a= R* alpha alpha=a/R?? acceleration = change in v / change in time

where can i get T1 or T2 since we have 2 equation and 3 unknow T1,T2 and a

T1-M1*g=M1*a how we use time to find acceleration?? since we have the giving acceleration, time: this is where i have no idea >.<

T1=35.8875N T1*R=1.7226 T2=-25.839 T2*R=-1.24027 I=0.000864 alpha=4.79167 i use all that and got 2.95873mN

... i got 2.95873mN for friction ... still wrong do we use time for some thing?

T1=M1*g-M1*a T2=M2*a+M2*g this is right? do i use the same "a" for each one?

Homework Statement A string passing over a pulley has a 3.75 kg mass hanging from one end and a 2.70 kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.8 cm and mass 0.75 kg. In fact, it is found that if the heavier mass is given a downward speed of...