A rod of mass M = 116 g and length L = 49 cm can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 14 g, moving with speed V= 6 m/s, strikes the rod at angle A = 37º a distance D=L/4 from the end and sticks to the rod after the collision. (A) What is the total moment of inertia, 1, with respect to the hinge, of the rod-ball-system after the collision?
(B) Enter an expression for the angular speed w of the system immediately after the collision, in terms of m, V, D, 0,
(C) Calculate the rotational kinetic energy, in joules, of the system after the collision

Answers

Answer 1
Answer:

Final answer:

We calculate the total moment of inertia of the rod-ball system after the collision by adding the moment of inertia of the rod and the added contribution from the putty ball. With this, we find the post-collision angular speed using Conservation of Angular Momentum. The rotational kinetic energy is then determined from this angular speed.

Explanation:

To solve this problem, we first need to calculate the moment of inertia of the combined system of the rod and the putty. The moment of inertia of an object is given by its mass times the square of its distance from the axis of rotation. That gives us I = 1/3 ML2 + m(D + L/2)2

Next we use Conservation of Angular Momentum to find the post-collision angular speed (ω). The initial momentum (mVD) is equal to the final moment of inertia times the final angular speed, so ω = mVD / I.

Finally, we calculate the rotational kinetic energy, which is given by ½ I ω2.

Learn more about moment of inertia here:

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Answers

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This is test content of administrator. Please keep calm and do not respond to it, do not report it, and remove it. It will be removed automatically

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This is a test answer.