How Do You Calculate the Total Force on a Pendulum's Bearing at 60 Degrees?

[chelito19]

Homework Statement

A pendulum has a mass of 7.5 kg with a center of mass at G (radius=250mm) and hass a radius of gyration about the pivot O of 295 mm. If the pendulum is released from rest at theta=0º, determine the total force supported by the bearing at the instant when theta=60º. Friction in the bearing is considered negligible

Homework Equations

At-Wsin(theta)=mr(alpha)
An-Wcos(theta)=mr(omega)^2
rAt=mk^2(alpha)

k=radius of gyration
At=tangential acceleration
An=normal acceleration

The Attempt at a Solution

I´m stuck. I know there has to be some relation between the moment of inertia and the acceleration but I don´t know how to solve this.

 

[AvroArrow]

I don´t even know where to start. I know that I have to use the equations of motion but that's it.Any help would be great.Since there is no friction in the bearing, the only force supported by the bearing will be the centrifugal force created by the acceleration of the pendulum. The centrifugal force can be calculated using the equations of motion.At-Wsin(theta)=mr(alpha)An-Wcos(theta)=mr(omega)^2Where W is the weight of the pendulum, m is the mass, r is the radius of the pendulum and alpha and omega are the angular acceleration and angular velocity respectively. In addition, the centrifugal force can also be calculated using the equation:rAt=mk^2(alpha)Where K is the radius of gyration of the pendulum.Substituting the values given in the question into the above equations, we get:At-7.5 x 9.81 x sin(60deg) = 7.5 x 0.25 x (alpha)An-7.5 x 9.81 x cos(60deg) = 7.5 x 0.25 x (omega)^2rAt = 7.5 x 0.295^2 x (alpha)Solving for alpha, we get:alpha = 8.14 rad/s^2Now, substituting this value of alpha into the equation, we get:rAt = 7.5 x 0.295^2 x 8.14 = 15.9 NTherefore, the total force supported by the bearing at the instant when theta=60º is 15.9 N.

 

[nlightner]

I would approach this problem by first drawing a free body diagram of the pendulum at the given position (theta=60º). This will help visualize the forces acting on the pendulum and their directions.

Next, I would use the equations of motion for rotational motion and apply them to the pendulum. The first equation, At-Wsin(theta)=mr(alpha), represents the sum of the tangential forces acting on the pendulum. In this case, there are two forces: the tension force (T) and the weight force (W). The second equation, An-Wcos(theta)=mr(omega)^2, represents the sum of the normal forces acting on the pendulum. In this case, there is only one normal force, which is the force exerted by the bearing.

Since the pendulum is released from rest, the initial angular velocity (omega) is equal to zero. This means we can simplify the second equation to An-Wcos(theta)=0. We also know that the tension force is always tangent to the path of motion, so we can write T=mr(omega)^2.

By substituting T and An into the first equation and solving for the normal force, we can find the total force supported by the bearing at theta=60º. This force will be equal to the normal force (An), since the friction in the bearing is considered negligible.

In order to use the third equation, rAt=mk^2(alpha), we need to find the value of the tangential acceleration (At). This can be done by using the equation for centripetal acceleration, Ac=(omega)^2r. We know the radius of the pendulum (r) and the angular velocity (omega) at theta=60º, so we can calculate the tangential acceleration (At).

Finally, by substituting the values of the tangential acceleration and the radius of gyration (k) into the third equation, we can solve for the moment of inertia (I). This will give us a better understanding of the distribution of mass in the pendulum, and how it affects the total force supported by the bearing.

In conclusion, by using the equations of motion for rotational motion and applying them to the given problem, we can solve for the total force supported by the bearing at theta=60º and gain a better understanding of the pendulum's motion.