A uniform rod of length L and mass m initially at rest is struck by a horizontal force F0 a distance x below the pivot O. The rod is free to rotate about its pivot without friction. If F0 is the average value of the force and Δt is its duration.

(a) What is the speed of the center of mass of the rod just after the rod is struck? Express your answer in terms of F0, Δt, L, x and M.
(b) What is the horizontal component of the force exerted by the pivot on the rod at the instant when the rod is struck? Express your answer in terms of F0, x, and L.
c)Where is the point of impact x such that the horizontal component of the force exerted by the pivot on the rod is zero? Express your answer in terms of L.

Problem#62

http://www.physics.umn.edu/classes/2012/fall/Phys%201401V.001/downloads/169171-Ch09_ISM.pdf

To solve this problem, we need to apply the principles of rotational motion and Newton's laws of motion. Let's go step by step:

(a) To find the speed of the center of mass of the rod just after it is struck, we can use the principle of conservation of angular momentum.

The angular momentum before the strike is zero since the rod is at rest. The angular momentum after the strike can be calculated as the product of the moment of inertia (I) and the angular velocity (ω) of the rod:

Angular momentum before = 0
Angular momentum after = I * ω

The moment of inertia of a uniform rod rotating about one end is given by: I = (1/3) * M * L^2, where M is the mass of the rod.

For a uniform rod rotating about one end, the angular velocity can be related to the linear velocity (v) of the center of mass as: ω = v / L.

Applying conservation of angular momentum: 0 = (1/3) * M * L^2 * v / L.

Simplifying, we get: v = 0.

Therefore, the speed of the center of mass of the rod just after it is struck is zero.

(b) The horizontal component of the force exerted by the pivot on the rod can be calculated using Newton's second law.

The net torque about the pivot point O is provided by the force F0 applied at a distance x below the pivot. The torque can be calculated as: Torque = (F0 * x)

According to rotational motion, Torque = I * α, where α is the angular acceleration. For a uniform rod rotating about one end, α can be related to the linear acceleration (a) of the center of mass as: α = a / L.

Using torque equation: (F0 * x) = ((1/3) * M * L^2) * (a / L).

Simplifying, we get: (F0 * x) = (1/3) * M * L * a.

Rearranging, we have: a = (3 * F0 * x) / (M * L).

Therefore, the horizontal component of the force exerted by the pivot on the rod is given by: F_horizontal = M * a = 3 * F0 * x.

(c) To find the point of impact x such that the horizontal component of the force exerted by the pivot on the rod is zero, we can set F_horizontal = 0 and solve for x.

From part (b), we found that the horizontal component of the force is given by: F_horizontal = 3 * F0 * x.

Setting F_horizontal = 0, we get: 3 * F0 * x = 0.

From the equation, we can see that the horizontal component of the force will be zero when either F0 = 0 (no force is applied) or x = 0 (the point of impact is at the pivot O).

Therefore, the point of impact x such that the horizontal component of the force exerted by the pivot on the rod is zero is either when F0 = 0 or when x = 0.