A disk of mass 50 kg and radius 25 cm rotates around

Consider a disk with a mass of 50 kg and a radius of 25 cm, rotating around a fixed axis passing perpendicular to the plane of the disk through its center with an angular velocity of 8.0 rps.

A brake pad was pressed to the rim of the disc with a force of 40 N, under the influence of which the disc stopped after 10 s.

It is required to determine the friction coefficient.

Answer:

Write down the equation of motion of the disc:

Ia = MR²a = FR - fR,

Where I – moment of inertia of the disk,

a – angular acceleration of the disk,

M – disk mass,

R – disc radius,

F – force acting on the disk,

f – friction force between the disc and the brake pad.

The moment of inertia of the disk is calculated by the formula:

I = MR²/2.

The angular acceleration of the disk is defined as:

a = v/(R/2),

Where v – speed of the disk edge.

The speed of the edge of the disk is:

v = πRn,

Where n – number of revolutions per second.

Substituting the expressions for the moment of inertia, angular acceleration and velocity into the equation of motion, we obtain:

MR²a = FR - fR,

MR²/2(v/(R/2)²) = FR - fR,

f = F - MaR/2(v/(R/2)²).

Substituting numerical values, we get:

f = 40 - 50 * (8 * 2π/60)² * 0.25/2 = -3.49 N.

Since the friction force cannot be negative, the friction coefficient is equal to:

m = |f|/F = 3,49/40 = 0,087.

Answer: The coefficient of friction between the disc and the brake pad is 0.087.

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From the description of the problem it follows that we are talking about a physical object - a disk with a mass of 50 kg and a radius of 25 cm, rotating around a fixed axis passing through its center. The disk makes 8.0 revolutions per second. A brake pad was applied to the rim of the disc with a force of 40 N, causing the disc to stop after 10 seconds.

To solve the problem, the equation of disk motion is used, which takes into account the moment of inertia of the disk, angular acceleration and forces acting on the disk, including the friction force between the disk and the brake pad. The moment of inertia of the disk is calculated by the formula I = MR^2/2, and the angular acceleration is calculated as α = v/(R/2), where v is the speed of the edge of the disk. The speed of the disk edge is defined as v = πRn, where n is the number of revolutions per second.

By substituting numerical values ​​into the equation of motion, you can determine the friction force between the disc and the brake pad. Then, using the formula for the coefficient of friction μ = |f|/F, where f is the friction force and F is the force acting on the disk, the coefficient of friction can be determined.

Thus, the answer to the problem is to determine the coefficient of friction between the disc and the brake pad, which is 0.087.

From the description of the problem it follows that we are talking about solid body mechanics. Specifically, we consider a disk with a mass of 50 kg and a radius of 25 cm, which rotates around a fixed axis at 8.0 rps. A brake pad was pressed to the rim of the disc with a force of 40 N, under the influence of which the disc stopped after 10 s.

To solve the problem, it is necessary to determine the coefficient of friction between the disc and the brake pad. To do this, the equation of motion of the disk is used, which relates the moment of inertia of the disk, angular acceleration and forces acting on the disk. From the equation it follows that the friction force between the disc and the brake pad is equal to the difference in forces acting on the disc. Substituting numerical values, we find that the coefficient of friction between the disc and the brake pad is 0.087.

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A disk with a mass of 50 kg and a radius of 25 cm rotates around a fixed axis passing perpendicular to the plane of the disk through its center with an angular velocity of 8.0 rps (radians per second). A brake pad was pressed to the rim of the disc with a force of 40 N, under the influence of which the disc stopped after 10 s. It is necessary to determine the friction coefficient.

To solve the problem, we will use the law of conservation of angular momentum. Before braking the disk, the angular momentum is equal to the angular momentum after braking. The angular momentum of the disk is defined as the product of the moment of inertia and angular velocity:

L = Iω

where L is the angular momentum, I is the moment of inertia of the disk, ω is the angular velocity of the disk.

The moment of inertia of the disk is defined as half the product of the mass of the disk and the square of its radius:

I = 1/2mr^2

where m is the mass of the disk, r is the radius of the disk.

After applying the braking force, the disk begins to rotate with angular acceleration, which is defined as the ratio of the moment of force to the moment of inertia:

α = τ/I

where τ is the moment of force acting on the disk.

The braking force creates a moment of force equal to the product of its magnitude by the radius of the disk:

τ = Fr

where F is the friction force acting on the disk.

Thus, we can write the equation for the angular momentum after braking:

L = Iω' = Fr'(t - t0)

where ω' is the angular velocity of the disk after braking, r' is the radius of the disk on which the friction force acts, t0 is the start time of braking.

Let us express the friction force from this equation:

F = I(ω' - ω)/(r'(t - t0))

Let's substitute the known values:

m = 50 kg (disk mass)

r = 0.25 m (disk radius)

ω = 8.0 r/s = 50.24 rad/s (angular speed of the disk before braking)

t0 = 0 s (start time of braking)

t = 10 s (disk stop time)

F = 40 N (braking force value)

Let's substitute all the values ​​and find the friction coefficient:

μ = F/(I(ω' - ω)/(r'(t - t0))) = 0.21

Thus, the coefficient of friction between the disc and the brake pad is 0.21.

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