From the case we know that:
- The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
- The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
- The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².
Please refer to the image below.
We know from the case, that:
m = 2M
r = R
m2 = 1/2M
distance between the center of mass to point P = p = R
Distance of the point mass to point P = d = 2R
We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:
Icm = 1/2mr²
Icm = 1/2(2M)(R²)
Icm = MR² ... (i)
Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:
Ip = Icm + mp²
Ip = MR² + (2M)R²
Ip = 3MR² ... (ii)
Then, the total moment of inertia of the disk with the point mass is:
I total = Ip + I mass
I total = 3MR² + (1/2M)(2R)²
I total = 3MR² + 2MR²
I total = 5MR² ... (iii)
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Answer:
The answer to your question is vo = 5.43 m/s
Explanation:
Data
distance = d= 5.8 m
height = 3 m
height 2 = 1.7 m
angle = 60°
vo = ?
g = 9.81 m/s²
Formula
hmax = vo²sinФ/ 2g
Solve for vo²
vo² = 2ghmax / sinФ
Substitution
vo² = 2(9.81)(3 - 1.7) / 0.866
Simplification
vo² = 19.62(1.3) / 0.866
vo² = 25.51 / 0.866
vo² = 29.45
Result
vo = 5.43 m/s
Weight = mg, g ≈ 9.8 m/s²
Weight = 2.2 * 9.8 ≈ 21.56 N
That is called fully constructive interference.
Answer:
R = v^2 sin 2 theta / g
The range provides the distance a projectile can travel
R(max) = v^2 / g if theta = 45 deg
R = 2.7^2 / 9.8 = .74 m