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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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leonid [27]

Answer:

a)

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b)

a = 1.6 [m/s^2]

c)

xt = 2100 [m]

Explanation:

In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.

a)

When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:

v_{f} = v_{i}+(a*t)

where:

Vf = final velocity = 40 [m/s]

Vi = initial velocity = 0 (starting from rest)

a = acceleration [m/s^2]

t = time = 20 [s]

40 = 0 + (a*20)

a = 2 [m/s^2]

The distance can be calculates as follows:

v_{f} ^{2} =  v_{i} ^{2}+(2*a*x)

where:

x1 = distance [m]

40^2 = 0 + (2*2*x1)

x1 = 400 [m]

Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.

v = x2/t2

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x2 = distance [m]

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x2 = 40*30

x2 = 1200 [m]

b)

Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.

v_{f} = v_{i}-a*t

Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.

0 = 40 - (a *25)

a = 40/25

a = 1.6 [m/s^2]

The distance can be calculates as follows:

v_{f} ^{2}  = v_{i} ^{2} -2*a*x3\\

0 = (40^2) - (2*1.6*x3)

x3 = 500 [m]

c)

Now we sum all the distances calculated:

xt = x1 + x2 + x3

xt = 400 + 1200 + 500

xt = 2100 [m]

8 0
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A plane flies from base camp to lake a, 200 km away in the direction 20.0° north of east. after dropping off supplies it flies t
Liono4ka [1.6K]

Distance of lake a is 200 km at 20 degree north of east

distance between lake a and b is 230 km at 30 degree west of north

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d = d_1 + d_2

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d_2 = -115 i + 199.2 j

now the total distance is

d = (187.94 - 115)i + (199.2 + 68.4)j

d = 72.94 i + 267.6 j

now the magnitude of the distance is given as

d = \sqrt{72.94^2 + 267.6^2}

d = 277.4

also the direction is given as

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