In a long jump, an athlete leaves the ground with an initial angular momentum that tends to rotate his body forward, threatening to ruin his landing. To counter this tendency, he rotates his outstretched arms to "take up" the angular momentum. In 0.393 s, one arm sweeps through 0.959 rev and the other arm sweeps through 0.336 rev. Treat each arm as a thin rod of mass 4.0 kg and length 0.60 m, rotating around one end. In the athlete's reference frame, what is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

Question

In a long jump, an athlete leaves the ground with an initial angular momentum that tends to rotate his body forward, threatening to ruin his landing. To counter this tendency, he rotates his outstretched arms to "take up" the angular momentum. In 0.393 s, one arm sweeps through 0.959 rev and the other arm sweeps through 0.336 rev. Treat each arm as a thin rod of mass 4.0 kg and length 0.60 m, rotating around one end. In the athlete's reference frame, what is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

Answer 1

To find the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders, we need to calculate the individual angular momenta of each arm and then add them up.

The angular momentum of a rotating object can be calculated using the formula:

L = Iω

Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

For a thin rod rotating around one end, the moment of inertia can be calculated using the formula:

I = (1/3)ml^2

Where m is the mass of the rod and l is its length.

Let's calculate the angular momentum for each arm:

Arm 1:
Mass (m) = 4.0 kg
Length (l) = 0.60 m
Angular displacement (θ1) = 0.959 rev = 2π * 0.959 radians

Using the formula for moment of inertia, we have:
I1 = (1/3) * 4.0 kg * (0.60 m)^2

Using the formula for angular momentum, we have:
L1 = I1 * ω1

Arm 2:
Mass (m) = 4.0 kg
Length (l) = 0.60 m
Angular displacement (θ2) = 0.336 rev = 2π * 0.336 radians

Using the formula for moment of inertia, we have:
I2 = (1/3) * 4.0 kg * (0.60 m)^2

Using the formula for angular momentum, we have:
L2 = I2 * ω2

To find the angular velocities (ω1 and ω2), we need to calculate the time it takes for each arm to sweep through their respective angular displacements.

Time (t) = 0.393 s

Using the formula for angular velocity, we have:
ω1 = θ1 / t
ω2 = θ2 / t

Finally, we can calculate the total angular momentum (L) by adding up the individual angular momenta (L1 and L2):

L = L1 + L2

Once you have the values for L, you can calculate its magnitude by taking the absolute value of L.

Answer 2

Figure I for each arm, as a thin rod rotating at the end.

Then angular mmentum=I1*angle1/time1+ I2*angle2/time2