Recent content by JLPG

It's mechanics, I'm guessing around grade 12, but I'm from Quebec, so here it's CEGEP. It's a crap system, but anyways, here is what I know from dynamics -the 3 basic equations for rotational kinematics -Torque=I(alpha) I=mr^2 for a point on a mass I know how to calculate I for a continuous rod...

Ok, but the answer is still wrong and I don't know where my mistake is...

I'm so sorry! I got the angle theta to be 115.322deg 1.35(9.8)(.355)sin115.322=4.2453 alpha=4.2453/.31=13.6948 a=13.6948*.83=11.3666 I know it's wrong because tangential acceleration should be greater than the one found in part a)

sin90/0.83=sin(theta)/0.355 sin-1= 23.8deg then, I followed the same steps as I did for part a). Torque=1.35*9.8*.355*sin23.8 Inertia is the same alpha=T/I

I'm still not getting the right answer, is the center of mass located at the center or is it more to the left now?

So, for b), moment of Inertia would be the same? Then for the torque=Frsin(theta), do I find the vertical component of gravity?

It worked! Thank-you so much!, I'm going to try b) now..

Torque=13.23(.83/2)=5.49N*m alpha= 5.49/.31=17.71rad/s^2 a(tangental)= 17.71*.83/2=7.35m/s^2 Where did I go wrong again?

Okay, but how do I apply this to the problem^ Do I use the length 0.83/2?

Center mass?

Ohhh, do I have to use the length of the string?

83 cm converted to .83 m... also tries using half the mass because each rope held 1/2 the weight of the rod, it didn't work either

Homework Statement A solid rod of mass M = 1.35 kg and length L = 83 cm is suspended by two strings, each with a length d = 71 cm (see Figure), one at each end of the rod. The string on side B is cut. What is the magnitude of the initial tangential acceleration of end B...

Could someone help me please?

http://gauss.vaniercollege.qc.ca/webwork2_course_files/NYAzhang/tmp/gif/1262751-1315-setDynamics_no_friction-prob14--prob56a.gif Sorry