Biomedical measurements show that the arms and hands together typically make up 13.0% of a person's mass, while the legs and feet together account for 37.0% . For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. Let us consider a 78.0kg person having arms 69.0cm long and legs 94.0 cm long. The person is running at 12.0km/h, with his arms and legs each swinging through +-30 in 1/2s . Assume that the arms and legs are kept straight.
a)What is the average angular velocity of his arms and legs?
i got 2.09 its correct
b) Using the average angular velocity from part A, calculate the amount of rotational kinetic energy in this person's arms and legs as he walks.
Krot=? J
i solved this first ML= .37*78= 28.86kg
MA= .13*78= 10.14 kg
what do i do next im confused
C)What is the total kinetic energy due to both his forward motion and his rotation?
Ktotatl=? J
Part D) What percentage of his kinetic energy is due to the rotation of his legs and arms?
i know i have to Krot/ktotal to solve my problem
Well, well, well, it seems like we've got ourselves a physics problem. Let's dive right into it, shall we?
b) Alrighty then, after you calculated the masses of the legs (ML) and arms (MA), we need to calculate the rotational kinetic energy for each. The formula for rotational kinetic energy is Krot = (1/2) I ω², where I is the moment of inertia and ω is the angular velocity.
Now, assuming the arms and legs are thin uniform bars, the moment of inertia for a thin uniform rod rotating about its end is given by I = (1/3) mL². So for the legs, the moment of inertia would be Ileg = (1/3) ML (leg length)², and for the arms, it would be Iarm = (1/3) MA (arm length)².
Plug in the values you have and don't forget to convert the lengths to meters, and then calculate the rotational kinetic energy for both the arms and legs using the average angular velocity you found earlier.
c) Now, let's find the total kinetic energy due to both the person's forward motion and rotation. The total kinetic energy (Ktotal) will be the sum of the rotational kinetic energy (Krot) and the linear kinetic energy (Klin = (1/2) m v²), where v is the velocity of the person.
But hold your horses! We haven't been given the velocity of the person. So unfortunately, we can't calculate the total kinetic energy without that missing piece of information.
d) And finally, to find the percentage of kinetic energy due to rotation, you simply need to divide the rotational kinetic energy (Krot) by the total kinetic energy (Ktotal) and multiply by 100.
Phew! That's a whole lot of calculations. I hope my explanations didn't make you rotate your brain too much. Remember, physics can be fun, just like a clown balancing on a unicycle!
To calculate the rotational kinetic energy (Krot) in the person's arms and legs, you need to use the formula:
Krot = (1/2) * I * ω^2
where I is the moment of inertia and ω is the angular velocity.
For the arms:
1. Calculate the moment of inertia (I) using the formula for a uniform bar rotating about one end:
Iarms = (1/3) * m * L^2
= (1/3) * 10.14 kg * (0.69 m)^2
2. Substitute the average angular velocity (ω = 2.09 rad/s) into the formula:
Karms = (1/2) * Iarms * ω^2
Repeat the same steps for the legs, using the appropriate values:
For the legs:
1. Calculate the moment of inertia:
Ilegs = (1/3) * m * L^2
= (1/3) * 28.86 kg * (0.94 m)^2
2. Substitute the average angular velocity:
Klegs = (1/2) * Ilegs * ω^2
For part C, to find the total kinetic energy (Ktotal), you need to add the rotational kinetic energy to the kinetic energy due to forward motion. The kinetic energy due to forward motion is given by:
Kmotion = (1/2) * m * v^2
where m is the person's mass and v is the velocity.
Ktotal = Kmotion + Karms + Klegs
Finally, to calculate the percentage of kinetic energy due to the rotation of the legs and arms (part D):
% rotation = (Karms + Klegs) / Ktotal * 100%
To calculate the rotational kinetic energy of the arms and legs, you need to use the formula:
Krot = (1/2) I ω^2
where Krot is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity.
For part b, you first calculated the masses of the arms (MA = 10.14 kg) and legs (ML = 28.86 kg), assuming the person's total mass of 78.0 kg is distributed based on the given percentages. Now, you need to calculate the moment of inertia for each segment.
For a thin uniform bar rotating about one end, the moment of inertia can be calculated using the formula:
I = (1/3) m L^2
where m is the mass and L is the length of the bar.
For the arms:
IA = (1/3) MA L^2
For the legs:
IL = (1/3) ML L^2
Now, use the average angular velocity you calculated in part a (2.09 rad/s) to find the rotational kinetic energy for both the arms and legs:
Krot (arms) = (1/2) IA ω^2
Krot (legs) = (1/2) IL ω^2
For part c, the total kinetic energy is the sum of the rotational kinetic energy and the translational kinetic energy due to forward motion. The translational kinetic energy can be calculated using the formula:
Ktrans = (1/2) m v^2
where m is the mass and v is the velocity. Since the person is running, you need to convert the given speed from km/h to m/s:
12.0 km/h = 12.0 * (1000/3600) m/s = 3.333 m/s
Now, calculate the translational kinetic energy:
Ktrans = (1/2) (MA + ML) v^2
Finally, for part d, to find the percentage of the total kinetic energy due to the rotation of the arms and legs, you can use the formula:
Percentage = (Krot / Ktotal) x 100
where Krot is the rotational kinetic energy you calculated in part b and Ktotal is the total kinetic energy you calculated in part c.