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2 years ago

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Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more m

ass, but has hexagonal sections where material has been removed. The attached masses are released from rest and allowed to fall a height h.Which of the following statements about their angular accelerations is true? a. The angular acceleration of the two flywheels is different but it is impossible to tell which is greater. b. The angular acceleration of flywheel A is greater The angular acceleration of flywheel B is greater. c. Not enough information is provided to determine. d. The angular accelerations of the two flywheels are equal.

1 answer:

inysia [295]2 years ago

4 0

Answer:

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

τ = I α

T r = I α

T gR = I α

Alf = T R / I (1)

T = α I / R

Now let's use Newton's second law in the mass that descends

W- T = m a

a = (m g -T) / m

The two accelerations need related

a = R α

α = a / R

a = (m g - α I / R) / m

R α = g - α I /m R

α (R + I / mR) = g

α = g / R (1 + I / mR²)

We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

Let's review the claims

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases

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5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by

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Answer:

a) W=2.425kJ

b) $\Delta E=2.425kJ$

c) $T_f=20.06^{o}C$

d) Q=-2.425kJ

Explanation:

a)

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

$W=Fd$

Where:

W=work done [J]

F= force applied [N]

d= distance [m]

In this case since it will be a vertical movement, the force is calculated like this:

F=mg

and the distance will be the height

d=h

so the formula gets the following shape:

$W=mgh$

so now e can substitute:

$W=(25kg)(9.7 m/s^{2})(10m)$

which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

$\Delta E=2.425kJ$

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

$\Delta Q=mC_{p}(T_{f}-T_{0})$

Where:

Q= heat transferred

m=mass

$C_{p}$ =specific heat

$T_{f}$ = Final temperature.

$T_{0}$ = initial temperature.

So we can solve the forula for the final temperature so we get:

$T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}$

So now we can substitute the data we know:

$T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C$

Which yields:

$T_{f}=20.06^{o}C$

d)

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

$\Delta Q=-2.425kJ$

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Answer:

8.67807 N

34.7123 N

Explanation:

m = Mass of shark = 92 kg

$\rho_{se}$ = Density of seawater = 1030 kg/m³

$\rho_{f}$ = Density of freshwater = 1000 kg/m³

$\rho_{sh}$ = Density of shark = 1040 kg/m³

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Net force on the fin is (seawater)

$F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N$

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Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

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