A uniform rod m1 and having a length L is hanging vertically downward and pin supported at the center of mass of a roller (disc) which rides on a horizontal track. Assuming that the system was initially at rest find the initial acceleration of the roller and angular acceleration of the rod. The roller's mass is m2, a force F1 (located at the pin of the rod and the roller) and a force F2 (located at the tip of the rod) both act horizontally to the right. The radius of the roller is R. Note that gravity acts, and the force vectors remain horizontal.
Choose x to be positive to the right and θ to be positive in the counterclockwise direction.
Find double dot theta and double dot x, both at t=0.
To find the initial acceleration of the roller (double dot x) and the angular acceleration of the rod (double dot theta) at time t=0, we need to analyze the forces acting on the system.
Let's consider the forces acting on the roller first. The only horizontal forces acting on the roller are F1 and F2. There are no vertical forces since gravity acts vertically downward. Since the system is initially at rest, the net force on the roller in the horizontal direction must be zero.
On the other hand, for the rod's angular acceleration, we need to consider the torque acting on the rod. The torque is given by the product of the force and the lever arm (perpendicular distance from the force to the axis of rotation).
In this case, the torque due to force F1 is zero since its lever arm is zero (it acts at the pin, which is the axis of rotation). The torque due to force F2 can be calculated as F2 multiplied by the lever arm, which is half the length of the rod (L/2).
Now, let's set up the equations for the forces and torques:
Net force on the roller (roller's mass m2): F1 + F2 - friction (assume there is no friction) = 0. Therefore, F1 + F2 = 0.
Torque on the rod (rod's mass m1): Torque = F2 * (L/2) = (m1 * R^2 * double dot theta) / 4 (using the parallel axis theorem).
Now, considering the given information that F1 and F2 act horizontally, we can rewrite the equations as:
F1 = -F2 (since they have opposite directions).
F1 + F2 = 0.
Simplifying the equations, we have:
2F1 = 0, which implies F1 = 0.
Now, if F1 = 0, the net force on the roller is zero. Therefore, the normal force and gravitational force must balance each other. The normal force equals the roller's weight, so the roller's weight is mg = m2 * g.
Since acceleration is defined as the net force divided by mass, the initial acceleration of the roller, double dot x at t=0, is:
double dot x = (F1 + F2) / m2 = 0 / m2 = 0.
For the angular acceleration of the rod, double dot theta at t=0, we substitute F1 = 0 into the torque equation:
F2 * (L/2) = (m1 * R^2 * double dot theta) / 4.
Since F2 is an external applied force acting on the rod, we need more information to calculate its value in order to find the angular acceleration at t=0.
In summary, based on the given information and conditions, the initial acceleration of the roller, double dot x at t=0, is 0, and the angular acceleration of the rod, double dot theta at t=0, cannot be determined without the value of force F2.