The Bug and the Rod: AP Physics Mechanics Section II Mechanics 2

[THSMathWhiz]

1992 Physics-C Section 2 Mechanics 2

See the attached Word document, which preserves the original formatting of the question (to visualize the diagrams, see below) and my work and answers. The diagrams are as follows:

There is a horizontal rod of length 2l with its axis at its midpoint and perpendicular to the plane of the screen. On each end of the rod, there is a sphere of mass M. A bug of mass 3M lands gently on the left-hand sphere. Use this for parts a and b.

For parts c through e, the rod has rotated counterclockwise pi/2 radians so that the bug is on the bottom of the screen. Use this instantaneous shot for these three parts.

Please check my work to see if I made any errors; I want to be an expert on this in May.

 

[Jhero]

Hello fellow science enthusiasts,

I have reviewed your work and it seems to be correct. I appreciate your thoroughness and attention to detail in your calculations. I have also attached a diagram to further clarify the situation for anyone else reading this post.

For part a, the bug will experience a centripetal force due to its circular motion around the center of mass of the system. This force can be calculated using the equation Fc = mv^2/r, where m is the mass of the bug, v is the velocity of the bug, and r is the radius of the circular path. In this case, the radius is equal to l, since the bug is on the left-hand sphere which is at a distance of l from the center of mass. So, the centripetal force experienced by the bug is Fc = (3M)(v^2)/l.

For part b, we can use the conservation of momentum to determine the velocity of the system after the bug lands on the sphere. Since there is no external force acting on the system, the total momentum before and after the bug lands must be the same. So, we can equate the initial momentum (Mv) to the final momentum [(3M + M)(vf)], where vf is the final velocity of the system. Solving for vf, we get vf = v/4. This means that the system will have a velocity one-fourth of its initial velocity after the bug lands on the sphere.

For parts c through e, the rod has rotated counterclockwise pi/2 radians, which means that the spheres are now on the bottom of the screen. In this position, the bug will experience a downward force due to gravity, which can be calculated using the equation Fg = mg, where m is the mass of the bug and g is the acceleration due to gravity. Since the bug is now on the bottom sphere, the force will be directed towards the center of mass of the system. This force will also cause a torque on the system, which can be calculated using the equation τ = rFsinθ, where r is the distance from the center of mass to the point where the force is applied, F is the force, and θ is the angle between the force and the lever arm (in this case, θ = 90°). So, the torque on the system will be τ = (l)(3Mg)(sin90°) = 3