# Rod and Pin Apparatus: Friction, Acceleration, and Velocity

Rigid rod OA is 2 ft long and rotated around fixed pt O. This rod is pinned at A to rigid connecting rod AB, 6 ft long, which in turn is pinned at B to piston C. This piston moves without friction along the y axis. The pins are also frictionless.

Rod OA is positioned so that it is pi/3 radians above the x-axis and angular velocity = 2 rad/sec counterclockwise and its angular acceleration is 0. What are the velocity and acceleration of the piston in this case?
and...
Weights of the three components are as follows: W(OA)=64.4 lb, W(AB) = 16.1 lb, W(C) = 5 lb. A constant counterclockwise torque is applied to rod OA at O. The system starts from rest with rod OA horizontal to the right. What torque is needed to give rod OA an angular velocity of 5 rad/sec when rod OA is vertical, along the y axis?

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#### Solution Preview

...ubstitute t = 0 to find the velocity, and we get:
Vt=0 = OA *  * cos /3 + 2* OA2 *  * (cos /3) * (sin /3) / (AB 2 - {OA* cos /3 }2)1/2 =
= 2*2*1/2 + 2*22*2*1/2*0.866/(62-(2*1/2)2)1/2 = 3.17 feet/s = 0.97 m/s.
Now, to find the acceleration, we have to find the derivative of velocity V over the time:
A = dV/dt = OA*(-sin  * d/dt *d/dt + cos  * d2/d2t ) +
+ OA2 * cos 2 * 2 * d/dt * d/dt / (AB 2 - {OA* cos }2)1/2 +
+ OA2 * sin 2 * d2/d2t / (AB 2 - {OA* cos }2)1/2 +
+ OA2 * sin 2 * d/dt * (-1/2) * ((-OA2)*2*cos  * (-sin ) * d/dt) / (AB 2 - {OA* cos }2)3/2
Since angluar acceleration of rod OA, d2/d2t = 0, we can simplify it:
A = - OA**sin  + 2*OA2 * 2 *cos 2 / (AB 2 - {OA* cos }2)1/2 -
- OA4 * sin 2 * 2 * cos  * sin  / (AB 2 - {OA* cos }2)3/2
Now, we substitute  = /3 and we get:
At=0 = -2*2*0.866 +2*4*4*(-1/2)/(36 - 1)1/2 - ...