Recent content by nopistons93

Answer: a) F= mgsin(theta) - ma b) Use T=rF and solve for T , then plug T into T=I*alpha and solve for I using a/r to find alpha. c) alpha = r*angular vel^2 , solve for ang vel and your done

anyone to confirm?

Yes, just putting radius in the formula gives you alpha. I do understand I think I'm just writing down my algebra wrong.

sorry for part b I got I=0.0958

well my equation yields 25.42 while the correct one yields 1.69

hey crypto, im at the point.. 2*a*d*sin(theta)=4gh/3 that reduces to a = 2g/3sin(theta) for some reason..

ok exactly what I was looking for there man. thank you very much\

does anyone know about this problem? for a) what are the external forces? I said F=ma = (12.0 kg)(cos 37)(2.0m/s) = 19.17 N b) I said Torque = radius*force = (0.1 m)(12 kg*cos(37)*2.00m/s) = 1.92 N*m thus solving for in \tau=I\alpha , I calculated 0.0958 c) I've calculated 14.1 m/s from...

im not understanding whre the Vcm^2 = 4gh/3 is coming from though..

ok so deriving the formula a = g*sin(theta)/1.5 , I found out that a = 1.69 m/s^2 since alpha = a/r , I found alpha to be 6.76 m/s^2. since alpha=r*omega^s , i found omega to be 5.20 m/s again. Solving for v from (v=r*omega), I found v equal to 1.30 m/s i then plugged these numbers into...

Homework Statement A solid uniform disk of radius 0.250 m and mass 55.0 kg rolls down an incline that makes an angle of 15 degrees with the horizontal. The disk starts from rest from the top of the ramp. Find the speed of the disk's center of mass when it reaches the bottom of the ramp(linear...