Question

A 7.1 kg, 5.0 m board is balanced about a point 2.0 m from right end of the board. A cube is sitting centered 0.5 m from the right end. A 3.5 kg cylinder is dropped from a height of 2.0 m onto the left end of the board and sticks to it. The dropped cylinder is going 6.26 m/s just before it hits the very left end of the board. (Assume the cube and cylinder are point masses.)

a) What is the mass of the cube sitting 0.5 m from the right end?
b) What is the cylinder’s angular momentum about the board/cube's pivot point just before
hitting the board?
c) What is the total angular momentum of all of the masses just after the collision?
d) What is the total moment of inertia of the board and two masses just after collision?
e) What is the board's angular velocity just after collision?
f) How fast is the cube (originally 0.5 m from the right end) going just after the collision?

This is due tomorrow night, so take your time, but please make sure that it is done well, I want to understand how you got your answers.

Answer 1

Step-by-step explanation:

To solve this problem, we can use the following equations:

Linear momentum: p = mv

Angular momentum: L = r x p

Total angular momentum: Ltotal = L1 + L2 + ... + Ln

Moment of inertia: I = mr^2

Angular velocity: w = L/I

Linear velocity: v = w x r

a) To find the mass of the cube sitting 0.5 m from the right end, we can use the equation p = mv. We know that p is the linear momentum of the cylinder just before hitting the board (which is equal to the mass of the cylinder times its velocity). We also know that the mass of the cylinder is 3.5 kg and its velocity is 6.26 m/s. Plugging these values into the equation gives us:

p = (3.5 kg)(6.26 m/s) = 22.01 kg m/s

We can use this value for the linear momentum of the cylinder to find the mass of the cube. Since the board is balanced about a point 2.0 m from the right end, the cube is 0.5 m from the pivot point and the cylinder is 2.5 m from the pivot point. This means that the cube and the cylinder have equal and opposite linear momentum about the pivot point. Therefore, the mass of the cube is equal to the mass of the cylinder, which is 3.5 kg.

b) To find the angular momentum of the cylinder just before hitting the board, we can use the equation L = r x p. We know that r is the distance from the pivot point to the center of mass of the cylinder (which is 2.5 m) and p is the linear momentum of the cylinder (which we just found to be 22.01 kg m/s). Plugging these values into the equation gives us:

L = (2.5 m) x (22.01 kg m/s) = 55.03 kg m^2/s

This is the angular momentum of the cylinder just before hitting the board.

c) To find the total angular momentum just after the collision, we need to find the angular momentum of the board and add it to the angular momentum of the cylinder. We can use the equation Ltotal = L1 + L2 + ... + Ln to do this. Since the board is balanced about a point 2.0 m from the right end, its angular momentum is equal to the mass of the board times the distance from the pivot point to the center of mass of the board times the velocity of the board. We know that the mass of the board is 7.1 kg, the distance from the pivot point to the center of mass of the board is 2.5 m, and the velocity of the board is 6.26 m/s (which is the same as the velocity of the cylinder just before hitting the board). Plugging these values into the equation gives us:

Ltotal = (7.1 kg)(2.5 m)(6.26 m/s) + 55.03 kg m^2/s = 122.64 kg m^2/s

This is the total angular momentum just after the collision.

d) To find the total moment of inertia just after the collision, we need to find the moment of inertia of the board and add it to the moment of inertia of the cylinder. We can use the equation I = mr^2 to find the moment of inertia of each object. For the board, we know that m is the mass of the board (which is 7.1