You should just ask the wave
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:
Angular momentum L = Iω, where
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>
<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>
<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>
<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>
<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
Answer:
1.2 rad/s
Explanation:
m1 = 15 g, m2 = 9 g, ω1 = 0.75 rad/s
Let the new angular speed is ω2 and the radius of the table be r.
The angular momentum is conserved when no external torque is applied.
I1 ω1 = I2 ω2
(m1 + m2)x r^2 x 0.75 = m1 x r^2 x ω2
(15 + 9) x 0.75 = 15 x ω2
ω2 = 1.2 rad/s
C because the wave length was longer
The force on the object has a constant strength, but its direction
keeps changing. The force is always directed from the object to
the center of the circle. It's called "centripetal force".