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liberstina [14]
2 years ago
14

Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th

e disk 2M 2 M Part (a) Calculate the moment of inertia ICM of the disk (without the point mass) with respect t0 the central axis of the disk in terms of M and R. Expression IcM Select from the variables below t0 write your expression. Note that all variables may not be required. 4,0, 0,4,d,gh,ijk M,P,R,$, Part (b) Calculate the moment of inertia [p of the disk (without the point mass) with respect to point P in terms of M and R. Expression Select from the variables below to write your expression_ Note that all variables may not be required: a,B,0,a,d,gh,ij,k M,PRS Part (c) Calculate the total moment of inertia [rof the disk with the point mass with respect to point P in terms of M and R Expression It Select from the variables below t0 write your expression Note that all variables may not be required: 4,0, 0,#,d,gh,ij,k M,P,R,$,'
Physics
1 answer:
brilliants [131]2 years ago
8 0

From the case we know that:

  1. The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
  2. The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
  3. 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

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The correct option that can be deduced for both Object P and Q is Option b) I and II only

To solve this question correctly, we need to understand the concept of density and it relation to mass and volume.

<h3>What is Density?</h3>

Density is a physical property of an object and can be expressed by using the relation:

\mathbf{Density = \dfrac{mass}{volume}}

From the given parameters, we are being told that:

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This implies that Q has a greater density that P. Since Q has a greater density than P, Q will be heavier since it will have greater mass.

However, Q will not be denser than water because if that happens, P will be have a greater density which is untrue in this scenario.

Therefore, we can conclude that:

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brainly.com/question/6838128

6 0
3 years ago
The heat capacity of nickel is 0.444 J/(g · °C). Calculate the amount of heat needed to raise the temperature of 18 g of nickel
azamat

The final speed of the nickel at the given quantity of heat is determined as 202.1  m/s.

<h3>Final speed of the nickel</h3>

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Q = 367.63 J

Q = K.E = ¹/₂mv²

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2 years ago
A small glass bead has been charged to +20 nC. A small metal ball bearing 1.0 cm above the bead feels a 0.018 N downward electri
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Answer:

q=1\times10^{-8}C

Explanation:

Let the charge on the ball bearing is q.

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Distance between them, d = 1 cm = 0.01 m

By use of Coulomb's law in electrostatics

F=\frac{KQq}{d^{2}}

By substituting the values

0.018=\frac{9\times10^{9}\times20\times10^{-9}q}{0.01^{2}}

q=1\times10^{-8}C

Thus, the charge on the ball bearing is q=1\times10^{-8}C

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This calculator is very helpful I use it on my homework

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