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SSSSS [86.1K]
2 years ago
10

Each of the gears a and b has a mass of 675 g and has a radius of gyration of 40 mm, while gear c has a mass of 3. 6 kg and a ra

dius of gyration of 100 mm. Assume that kinetic friction in the bearings of gears a, b, and c produces couples of constant magnitude 0. 15 n·m, 0. 15 n·m, and 0. 3 n·m, respectively. Knowing that the initial angular velocity of gear c is 2000 rpm, determine the time required for the system to come to rest.
Physics
1 answer:
navik [9.2K]2 years ago
6 0

9.87 seconds

The time required for this system to come to rest is equal to 9.87 seconds.

We have the following data:

Mass of gear A = 675 g to kg = 0.675 kg.

Radius of gear A = 40 mm to m = 0.04 m.

Mass of gear C = 3.6 kg.

Radius of gear C = 100 mm to m = 0.1 m.

How can I calculate the time needed?

We would need to figure out the moment of inertia for gears A and C in order to compute the time needed for this system to come to rest.

Mathematically, the following formula can be used to determine the moment of inertia for a gear:

I = mr²

Where:

m is the mass.

r is the radius.

We have, For gear A:

I = mr²

I = 0.675 × 0.04²

I = 0.675 × 0.0016

I = 1.08 × 10⁻³ kg·m².

We have, For gear C:

I = mr²

I = 3.6 × 0.1²

I = 3.6 × 0.01

I = 0.036 kg·m².

The initial angular velocity of gear C would therefore be converted as follows from rotations per minute (rpm) to radians per second (rad/s):

ωc₁ = 2000 × 2π/60

ωc₁ = 4000π/60

ωc₁ = 209.44 rad/s.

Also, the initial angular velocity of gears A and B is given by:

ωA₁ = ωB₁ = rc/rA × (ωc₁)

ωA₁ = ωB₁ = 0.15/0.06 × (209.44)

ωA₁ = ωB₁ = 2.5 × (209.44)

ωA₁ = ωB₁ = 523.60 rad/s.

Taking the moment about A, we have:

I_A·ωA₁ + rA∫F_{AC}dt - M(f)_A·t = 0

On Substituting the given parameters into the formula, we have;

(1.08 × 10⁻³)·(523.60) + 0.06∫F_{AC}dt - 0.15t = 0

0.15t - 0.06∫F_{AC}dt = 0.56549   ----->equation 1.

Similarly, the moment about B is given by:

0.15t - 0.06∫F_{BC}dt = 0.56549    ------>equation 2.

Note: Let x = ∫F_{BC}dt + ∫F_{AC}dt

Adding eqn. 1 & eqn. 2, we have:

0.3t - 0.06x = (0.56549) × 2

0.3t - 0.06x = 1.13098  ------>equation 3.

Taking the moment about A, we have:

Ic·ωc₁ - rC∫F_{AC}dt - rC∫F_{BC}dt - Mc(f)_A·t = 0

0.036(209.44) - 0.3t - 0.15(∫F_{BC}dt + ∫F_{AC}dt) = 0

0.3t + 0.15x = 7.5398    ------->equation 4.

Solving eqn. 3 and eqn. 4 simultaneously, we have:

x = 30.5 Ns.

Time, t = 9.87 seconds.

To learn more about moment of inertia visit:

brainly.com/question/15246709

#SPJ4

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1.)Answer:

The correct answer is C.

Explanation:

Displacement is a vector quantity so it may be either positive or negative. As per convention, acceleration taken as positive when moves downward direction. But, in our case, acceleration towards downward taken as negative, so displacement should also be negative because negative sign only indicate its downward direction.

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3.)Question

Illustration:

Day 1: 10 km west and spent night at A.

Day 2: 10 km west and night in a tent B.

Day 3: 20 km west and then they turned around and went back to the home because Jerry left his allergy medicine.

Day 4: 30 km North to the point D.

Answer

The correct answer is C.

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4.) Answer:

The correct Answer is A.

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The correct Answer is B.

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The correct answer is A.

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The correct answer is D.

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The correct Answer is B.

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The correct Answer is B.

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The correct answer is D.

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The correct answer is C.

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