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ICE Princess25 [194]
3 years ago
8

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

iform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar.
Physics
1 answer:
Kamila [148]3 years ago
3 0

Answer:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c)  </em>\frac{4492.18}{v_{car} ^{2} }

<em></em>

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

I = \frac{1}{2}mr^{2}

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

I =  \frac{1}{2}*11*1.1^{2} = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = Iw^{2} = 6.655 x 31.82^{2} = <em>6738.27 J</em>

<em></em>

b) second flywheel  has

radius = 2.8 m

mass = 16 kg

moment of inertia is

I = \frac{1}{2}mr^{2} =  \frac{1}{2}*16*2.8^{2} = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = Iw = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

(I_{1} +I_{2} )w

where the subscripts 1 and 2 indicates the values first and second  flywheels

(I_{1} +I_{2} )w = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = Iw^{2} = 6.655 x 3.05^{2} = <em>61.908 J</em>

<em></em>

<em></em>

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = \frac{1}{2}mv_{car} ^{2}

where m is the mass of the car

v_{car} is the velocity of the car

Equating the energy

2246.09 =  \frac{1}{2}mv_{car} ^{2}

making m the subject of the formula

mass of the car m = \frac{4492.18}{v_{car} ^{2} }

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A student sits on a pivoted stool while holding a pair of weights. The stool is free to rotate about a vertical axis with neglig
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Answer:

<u>Please Mark As Brainliest!!</u>

a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J

Explanation:

a)  If the student completes one turn in 1.26 sec, this is called the period of the movement.

If we take into account that the angle rotated during one turn is 2π rads, by definition of angular velocity, we can get this value as follows:

ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.

b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:

Li = Lf   ⇒  I₁ * ω₁  = I₂* ω₂

So, we can solve for ω₂, as follows:

ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec

c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:

W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²

W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)  

W = 7.03 J

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Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani
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Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

n = \frac{t}{t_{1/2}}

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

t_{1/2} = half-life = 700 million years

Therefore,

n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3

Now, we will calculate the number of uranium nuclei left (n_u):

n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u =  800\ million

and the rest of the uranium nuclei will become thorium nuclei (u_{th})

n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million

dividing both:

\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u

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6 0
2 years ago
Block A, with a mass of 4.0 kg, is moving with a speed of 2.0 m/s while block B , with a mass of 8.0 kg, is moving in the opposi
Anon25 [30]
Consider velocity to the right as positive.

First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right

Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left

Total momentum of the system is
P = m₁v₁ + m₂v₂
   = 4*2 + 8*(-3)
  = -16 (kg-m)/s

Let v (m/s) be the velocity of the center of mass of the 2-block system.

Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s

Answer:
The center of mass is moving at 1.33 m/s to the left.
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