A uniform rod is 2.0 m long. The rod is pivoted around a horizontal, frictionmuch less pin with one end. The rod is released from rest at an angle of 30° above the horizontal. What is the angular acceleration of the rod at the immediate it is released?

I simply supplied

\$\$sin(30) = frac9.8a_centripetal\$\$

Then I related

Is this right? I am looking at University rwandachamber.org:

Where 9.8 is gravity. But I obtained none of the answers in the multiple option ... so I have to be doing wrong. Also I haven"t offered the radius. Any suggestions would certainly be advantageous...

You are watching: What is the angular acceleration α of the rod immediately after it is released?

homework-and-exercises rotational-dynamics rotational-kinematics angular-velocity
Share
Cite
Improve this question
Follow
edited Oct 28 "12 at 12:23 Waffle's Crazy Peanut
asked Oct 28 "12 at 11:54 Jiew MengJiew Meng
\$endgroup\$
0

2

See more: Yandere Simulator How To Activate Easter Eggs In Yandere Simulator

\$egingroup\$
Almeans begin through a nice clear diagram/sketch of the problem. It all follows from there. Here is a Free Body Diagram I created you. Then you have actually (the long thorough way):

Sum of the forces on body amounts to mass times acceleration at the facility of gravity. \$sum_i vecF_i = m veca_C \$

\$\$ A_x = m a_x \ A_y - m g = m a_y \$\$

Sum of torques about center of gravity equates to minute of inertia times angular acceleration.\$sum_i left(vecM_i + (vecr_i-vecr_C) imesvecF_i ight) = I_C vecalpha \$

\$\$ A_x fracL2 sin( heta) - A_y fracL2 cos( heta) = I_C ddot heta \$\$

Acceleration of suggest A must be zero. \$veca_A = veca_C + vecalpha imes(vecr_A-vecr_C) + vecomega imes(vecv_A-vecv_C) \$

\$\$ a_x + fracL2 sin( heta) ddot heta + fracL2 dot heta^2 cos( heta) =0 \ a_y - fracL2 cos( heta) ddot heta + fracL2 dot heta^2 sin( heta) =0 \$\$

Now you deserve to settle for \$a_x\$, \$a_y\$ from 3. and also usage those in 1. to acquire \$A_x\$,\$A_y\$. Finally usage 2. to deal with for \$ddot heta\$

Or execute the shortcut of finding the used torque on A and also using it to the efficient minute of inertia about the pivot \$I_A = I_C + m left(fracL2 ight)^2 \$ to get

\$\$ ddot heta = fracm g fracL2 cos heta I_C + m left(fracL2 ight)^2 \$\$