In the figure, a 1.6-kg weight swings in a vertical circle at the end of a string having negligible weight. The string is 2 m long. If the weight is released with zero initial velocity from a horizontal position, its angular momentum (in kg⋅m²/s) at the lowest point of its path relative to the center of the circle is approximately A)40 B)10 C)30 D)20 E)50

Question

Quizplus · Accepted Answer

At the lowest point of the circle, the weight is at its minimum height, which is equal to the radius of the circle (2 m). At this point, both the gravitational force and the tension force of the string are acting on the weight, each having a component in the direction of the angular velocity. However, since the angular velocity is perpendicular to the plane of motion (the plane of the circle), its direction is perpendicular to the direction of the gravitational and tension forces.

Therefore, the angular momentum of the weight at the lowest point is equal to its momentum perpendicular to the plane of motion, which is simply the product of its linear momentum and the radius of the circle (angular momentum = momentum x radius). Since the weight is released with zero initial velocity, its linear momentum at the lowest point is equal to the momentum it gains due to gravity during its fall from the horizontal position.

The potential energy of the weight at the horizontal position is mgh, where m is its mass, g is the acceleration due to gravity, and h is the height of the weight relative to the lowest point of the circle. Since the weight is released from a horizontal position, h is equal to the length of the string (2 m). Therefore, the potential energy at the horizontal position is mgh = (1.6 kg)(9.81 m/s^2)(2 m) = 31.4 J.

At the lowest point of the circle, all of this potential energy is converted into kinetic energy. Therefore, the kinetic energy of the weight at the lowest point is also 31.4 J. The speed of the weight at the lowest point can be calculated from its kinetic energy:

KE = (1/2)mv^2
31.4 J = (1/2)(1.6 kg)v^2
v^2 = 39.125 m^2/s^2
v ≈ 6.25 m/s

The linear momentum of the weight at the lowest point is simply its mass times its velocity:

p = mv
p = (1.6 kg)(6.25 m/s)
p ≈ 10 kg·m/s

Finally, the angular momentum of the weight at the lowest point is the product of its linear momentum and the radius of the circle:

L = rp
L = (2 m)(10 kg·m/s)
L = 20 kg·m^2/s

Therefore, the angular momentum of the weight at the lowest point is approximately 20 kg·m^2/s, which is closest to choice D (20).

In the Figure, a 1