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)
Learn more about Uniform Flat Disk here: brainly.com/question/14595971
#SPJ4
Both the size and the shape of the tree changes
Answer:
230 m/s northeast, 1.8 m/s up
Explanation:
204 kilometres = 204000 metres
15.0 minutes = 900 seconds
Velocity = Distance / Time
= 204000 / 900
= 230 m/s northeast (to 2 sf.)
1.6km = 1600 metres
Velocity = 1600 / 900
= 1.8 m/s up (to 2 sf.)
Read more on Brainly.com - brainly.com/question/13863590#readmore
Answer:
890 N
Explanation:
Acceleration is change in velocity over change in time.
a = Δv / Δt
a = (11 m/s − 0 m/s) / 0.26 s
a = 42.3 m/s²
Force is mass times acceleration.
F = ma
F = (21 kg) (42.3 m/s²)
F ≈ 890 N
The players acceleration is 3.33 m/s/s
Acceleration= Velocity/Time
A =10/3