Question

A 75 g, 34-cm-long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 15 g ball of clay traveling horizontally at 2.2 m/s hits and sticks to the very bottom tip of the rod. What is the angular velocity of the rod-ball system after the collision?

1) 0.15 rad/s
2) 0.30 rad/s
3) 0.45 rad/s
4) 0.60 rad/s

Answer 1

The angular velocity of the rod-ball system after the collision is 0.045 rad/s.

Step-by-step explanation:

To find the angular velocity of the rod-ball system after the collision, we can use the principle of conservation of angular momentum. Before the collision, the rod is at rest, so its initial angular momentum is 0. After the collision, the clay ball sticks to the very bottom tip of the rod, which changes the moment of inertia of the system. The final angular momentum can be calculated using the equation:

Li + Lc = Lf

where Li is the initial angular momentum, Lc is the angular momentum of the clay ball, and Lf is the final angular momentum of the system. Since the clay ball is moving horizontally, its angular momentum is given by Lc = mc * vc * r, where mc is the mass of the clay ball, vc is its velocity, and r is the distance from the axis of rotation to the point where the clay ball sticks to the rod. Plugging in the known values, we can solve for the final angular velocity of the system.

Using the given values, we have:

Li + (mc * vc * r) = Lf

Since Li = 0, the equation simplifies to:

mc * vc * r = Lf

Plugging in the known values, we have:

(15g * 2.2m/s * 0.34m) = (75g + 15g) * Vf * r

Simplifying the equation gives:

1.122g * m/s = 90g * Vf * 0.34m

Solving for Vf gives:

Vf = 1.122g * m/s / (90g * 0.34m)

Therefore, Vf = 0.045 rad/s. So, the angular velocity of the rod-ball system after the collision is 0.045 rad/s.