uniform rod pivoted off-center

mercredi 31 juillet 2013

√ω1. The problem statement, all variables and given/known data

A uniform rod of mass 1.2 kg and length 1.2 m is pivoted in the horizontal position as shown (black point). The rod is at rest and then released. The acceleration due to gravity is g = 9.8 m/s².

Part 1. Find the rotational inertia of the rod relative to the axis perpendicular to the screen and passing through the pivot point (in kgm²).

Part 2. As the rod swings downward, its angular speed velocity and the magnitude of its angular acceleration, respectively are (increasing, decreasing, staying the same) and (increasing, decreasing, staying the same).

Part 3. What is the angular speed (in rad/s) of the rod as it passes through the vertical position (when end marked B is at the bottom)?

Part 4. What is the linear speed of the bottom end of the rod (marked B) when it passes the vertical position?(in m/s)

2. Relevant equations

I_cm=1/12ml²

I_p=I_cm+md²

U_g=mgh

U_i+K_i+W^noncon=U_f+K_f

K_trans=1/2mv²

K_rot=1/2Iω²

t=Frsinθ

K=K_trans+K_rot

v=ωR

3. The attempt at a solution

Picture/FBD is attached.

Part 1

I_p=1/12m(l)²+m(l/4)²=1/12*1.2*1.2²+1.2*.3²=.252

Part 2

increasing and decreasing

Part 3

0=1/2mv²+1/2Iω²+mgh where h=-l/4(distance from pivot to cm which will be directly below pivot when the rod is at the bottom of its swing)

mgh=1/2mv²+1/2Iω²=mgh=1/2mω²R²+1/2Iω²=1/2mω²R²+1/2Iω²

1.2(9.8)(.3)=(1/2(1.2)(.3)²+1/2*.252)ω²=(.18)ω²

ω=√(3.528/.18)=4.42718872424

I am not sure if this is correct, but I used l/4 (distance from pivot to center of mass) as R.

Part 4

mgh=1/2mv²+1/2Iω²

2mgh=mv²+Iv²/R²

2(1.2)(9.8)(.3)=(1.2+.252/.3²)v²

v=√(7.056/4)=1.32815661727

or v=ωR²=4.42718872424(.3)^2=0.398446985181

Since these two values for v don't match up, I think I'm doing something wrong.

Attached Images