can you tell me what i did wrong on this problem
Rigid rods of negligible mass lying along the y axis connect three particles. The system rotates about the x axis with an angular speed of 3.30 rad/s.
4kg,y=3.00m
2kg,y=-2.00m
3kg,y=-4.00m
a)Find the total rotational kinetic energy evaluated from 1/2 I omega^2
4(3^2)+2(2^2)+3(4^2)=92
.5(92)(3.3)^2
b)Find the tangential speed of each 4kg and 3kg particles
4kg: 4x3.3=13.2m/s
3kg: 3x3.3=9.9m/s
c)Find the total kinetic energy evaluated from Σ mi vi2/2
.5(4kg)(6^2)+.5(2kg)(4^2)+.5(3kg)(8^2)=184
i don't know why i got these wrong
Looking at your calculations, I can see that you made a mistake in part c) when finding the total kinetic energy evaluated from Σ mi vi^2/2.
Let's go through the calculations step by step:
a) To find the total rotational kinetic energy, you correctly used the formula 1/2 I ω^2, where I is the moment of inertia and ω is the angular speed.
You calculated it as follows:
4(3^2) + 2(2^2) + 3(4^2) = 92
0.5(92)(3.3)^2 = 523.38
So, the total rotational kinetic energy is 523.38 J. Your calculation is correct.
b) To find the tangential speed of each particle, you correctly multiplied the angular speed (ω) by the perpendicular distance (y) from the rotation axis to each particle.
For the 4kg particle:
4kg × 3.3 m/s = 13.2 m/s
For the 3kg particle:
3kg × 3.3 m/s = 9.9 m/s
So, the tangential speeds of the 4kg and 3kg particles are 13.2 m/s and 9.9 m/s, respectively. Your calculations are correct.
c) Now let's find the total kinetic energy evaluated from Σ mi vi^2/2. This formula calculates the translational kinetic energy of each particle and sums them up.
You used the formula correctly and set it up as follows:
0.5(4kg)(6^2) + 0.5(2kg)(4^2) + 0.5(3kg)(8^2) = 184
However, there was a mistake in your calculation. Let's correct it:
0.5(4kg)(6^2) + 0.5(2kg)(4^2) + 0.5(3kg)(8^2) = 0.5(4)(36) + 0.5(2)(16) + 0.5(3)(64)
= 72 + 16 + 96 = 184
So, the total kinetic energy evaluated from Σ mi vi^2/2 is indeed 184 J. Your calculation was correct, but there was likely an arithmetic error in your final calculation.
In summary, it appears that you made a calculation mistake in part c) when finding the total kinetic energy. Make sure to carefully double-check your arithmetic to avoid such errors.
a) Your calculation for the total rotational kinetic energy seems correct. The rotational kinetic energy is given by 1/2 I omega^2, where I is the moment of inertia and omega is the angular speed. You correctly calculated the moment of inertia for each particle and summed them up. Then, you multiplied the total moment of inertia by (3.3)^2 to find the rotational kinetic energy. Your answer of 92 is correct.
b) To find the tangential speed of each particle, you need to multiply the radius (distance from the axis of rotation) by the angular speed (omega). However, you made an error in multiplying the masses by the angular speed instead of multiplying the radii.
For the 4 kg particle:
Tangential speed = radius × angular speed = 3.0 m × 3.3 rad/s = 9.9 m/s
For the 3 kg particle:
Tangential speed = radius × angular speed = 4.0 m × 3.3 rad/s = 13.2 m/s
So, the correct tangential speeds are 9.9 m/s for the 4 kg particle and 13.2 m/s for the 3 kg particle.
c) The calculation for the total kinetic energy using Σ mi vi^2/2 seems incorrect. You squared the velocities instead of the tangential speeds, and you used different radii for each particle. Remember, the formula for kinetic energy is KE = 1/2 mv^2, where m is the mass and v is the speed.
For the 4 kg particle:
Kinetic energy = (1/2)(4 kg)(9.9 m/s)^2 = 196 J
For the 2 kg particle:
Kinetic energy = (1/2)(2 kg)(0 m/s)^2 = 0 J (as it is not moving)
For the 3 kg particle:
Kinetic energy = (1/2)(3 kg)(13.2 m/s)^2 = 274.68 J
Therefore, the total kinetic energy is 196 J (for the 4 kg particle) + 0 J (for the 2 kg particle) + 274.68 J (for the 3 kg particle) = 470.68 J.
Make sure to use the correct radii and tangential speeds when calculating kinetic energy.