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liberstina [14]
1 year 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]1 year 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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Answer:

 ∑ τ =0,  L₀ = L_{f}

Explanation:

In a circular turning movement, when the arms are extended and then contracted in two possibilities:

- They are lowered the force of gravity is what pulls them, the tension of the muscle becomes zero to allow this movement.

    In this movement the force is vertical(gravity) and the movement of the center of mass of each arm is vertical, so that the work is the weight value of the arm by the distance traveled by the center of mass.

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In these movements the torque of the external force is equal for each arm, but in the opposite direction, so they are canceled where a net torque of zero, this causes the angular momentum to be preserved, which changes is the moment of inertia of the system and therefore you must also change the angular velocity to keep your product constant

          ∑ τ =0

                L₀ = L_{f}

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4 0
3 years ago
A person of 70 kg standing on an un-deformable horizontal surface. She bends her knees and jumps up from rest, achieving a launc
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Answer:

1190 N

Explanation:

Force: This can be defined as the product of mass and velocity. The unit of force is Newton(N).

From the question,

F = ma................. Equation 1

Where F = average force, m = mass, a = acceleration.

But,

a = (v-u)/t................ Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

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Given: m = 70 kg, v = 1.7 m/s, u = 0 m/s (from rest), t = 0.1 s.

Substitute into equation 3

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Answer:

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Explanation:

We have given mass of the car m = 500 kg

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From second equation of motion

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0^2=14.5^2+2\times a\times 18.25

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We know that acceleration is given by

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