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ICE Princess25 [194]
2 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]2 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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Explanation:

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We need to find the width of the hair. Using the condition of diffraction pattern as :

s=\dfrac{m\lambda l}{d}, d is the width of the hair

d=\dfrac{m\lambda l}{s}

d=\dfrac{1\times 6\times 10^{-7}\times 12}{0.007}

d = 0.00102

or

d=1.02\times 10^{-3}\ m

So, the width of the hair is 1.02\times 10^{-3}\ m. Hence, this is the required solution.

8 0
2 years ago
A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards
8_murik_8 [283]

Answer:

<em>b. The current in the loop always flows in a counterclockwise direction.</em>

<em></em>

Explanation:

When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.

3 0
2 years ago
A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows:
Nikitich [7]

Answer:

r ≥ R, E = Q / (4πR²ε₀)

r ≤ R, E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

Maximum at r = ⅔ R

Maximum field of E = Q / (3πε₀R²)

Explanation:

Gauss's law states:

∮E·dA = Q/ε₀

What that means is, if you have electric field vectors E passing through areas dA, the sum of those E vector components perpendicular to the dA areas is equal to the total charge Q divided by the permittivity of space, ε₀.

a) r ≥ R

Here, we're looking at the charge contained by the entire sphere.  The surface area of the sphere is 4πR², and the charge it contains is Q.  Therefore:

E(4πR²) = Q/ε₀

E = Q / (4πR²ε₀)

b) r ≤ R

This time, we're looking at the charge contained by part of the sphere.

Imagine the sphere is actually an infinite number of shells, like Russian nesting dolls.  For any shell of radius r, the charge it contains is:

dq = ρ dV

dq = ρ (4πr²) dr

The total charge contained by the shells from 0 to r is:

q = ∫ dq

q = ∫₀ʳ ρ (4πr²) dr

q = ∫₀ʳ ρ₀ (1 − r/R) (4πr²) dr

q = 4πρ₀ ∫₀ʳ (1 − r/R) (r²) dr

q = 4πρ₀ ∫₀ʳ (r² − r³/R) dr

q = 4πρ₀ (⅓ r³ − ¼ r⁴/R) |₀ʳ

q = 4πρ₀ (⅓ r³ − ¼ r⁴/R)

Since ρ₀ = 3Q/(πR³):

q = 4π (3Q/(πR³)) (⅓ r³ − ¼ r⁴/R)

q = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴)

Therefore:

E(4πr²) = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / ε₀

E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

When E is a maximum, dE/dr is 0.

First, simplify E:

E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

E = Q (4 (r³/R³) − 3 (r⁴/R⁴)) / (4πr²ε₀)

E = Q (4 (r/R³) − 3 (r²/R⁴)) / (4πε₀)

Take derivative and set to 0:

dE/dr = Q (4/R³ − 6r/R⁴) / (4πε₀)

0 = Q (4/R³ − 6r/R⁴) / (4πε₀)

0 = 4/R³ − 6r/R⁴

0 = 4R − 6r

r = ⅔R

Evaluating E at r = ⅔R:

E = Q (4 (⅔R / R³) − 3 (⁴/₉R² / R⁴)) / (4πε₀)

E = Q (8 / (3R²) − 4 / (3R²)) / (4πε₀)

E = Q (4 / (3R²)) / (4πε₀)

E = Q (1 / (3R²)) / (πε₀)

E = Q / (3πε₀R²)

3 0
1 year ago
Dźwig podniósł kontener o masie m = 80 kg na wysokość h = 10 m. Pierwsze 5 m kontener przebył z przy-
Nady [450]

Answer:

a)   W_total = 8240 J , b)  W₁ / W₂ = 1.1

Explanation:

In this exercise you are asked to calculate the work that is defined by

       W = F. dy

As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.

       W = F dy = F Δy

let's apply this formula to our case

a) Let's use Newton's second law to calculate the force in the first y = 5 m

          F - W = m a

          W = mg

          F = m (a + g)

          F = 80 (1 + 9.8)

          F = 864 N

The work of this force we will call it W1

   We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)

        F₂ - W = 0

        F₂ = W

        F₂ = 80 9.8

        F₂ = 784 N

The work of this fura we will call them W2

The total work is

         W_total = W₁ + W₂

         W_total = (F + F₂) y

         W_total = (864 + 784) 5

         W_total = 8240 J

b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use

        W₁ / W₂ = F y / F₂ y

        W₁ / W₂ = 864/784

        W₁ / W₂ = 1.1

7 0
2 years ago
Which statement best explains why hydrogen’s atomic number is equal to its mass?(1 point)
igor_vitrenko [27]

Answer:

Hydrogen has one electron and one proton

3 0
1 year ago
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