The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.
1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.
a. the same as before.
b. lower than before.
c. higher than before.
3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?
4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?
4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.
A. Only the odd multiples of the fundamental frequency.
B. Only the even multiples of the fundamental frequency.
C. All integer multiples of the fundamental frequency.
E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?
A. Half the length of the open-open pipe.
B. Twice the length of the open-open pipe.
C. One-fourth the length of the open-open pipe.
D. Four times the length of the open-open pipe.
E. The same as the length of the open-open pipe.
F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third
harmonic.

Answer:

A)     K / K₀ = 4   b)     v / v₀ = 4

Explanation:

A) For this exercise we can use the conservation of mechanical energy

in the problem it indicates that the displacement was doubled (x = 2xo)

starting point. At the position of maximum displacement

      Em₀ = Ke = ½ k (2x₀)²

final point. In the equilibrium position

      [tex]Em_{f}[/tex] = K = ½ m v²

        Em₀ = Em_{f}

        ½ k 4 x₀² = K

        (½ K x₀²) = K₀

         K = 4 K₀

          K / K₀ = 4

B) the speed value

          ½ k 4 x₀² = ½ m v²

          v = 4 (k / m) x₀

if we call

           v₀ = k / m x₀

          v = 4 v₀

         v / v₀ = 4

Answer:

The potential will be Va/b

Explanation:

So Let sphere A charged Q to potential V.

so, V= KQ/a. ....(1

Thus, spherical shell B is connected to the sphere A by a wire, so all charge always reside on the outer surface.

therefore, potential will be ,

V ′ = KQ/b = Va/b... That is from .....(1), KQ=Va]

Answer:

Explanation:

The e.m.f induced in the coil depend on the following :

(a) No. of turns in the coil

(b) Cross-sectional Area of the coil

(c) Magnitude of Magnetic field

(d) Angular velocity of the coil

Answer:

[tex]\theta=10.20^{\circ}[/tex]  

[tex]\Delta l=0.10 ft[/tex]    

Explanation:

First of all, we analyze the system of blocks before starting to move.

[tex]\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0[/tex]  

[tex]\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]  

[tex]16sin(\theta)-2.91cos(\theta)=0[/tex]  

[tex]tan(\theta)=0.18[/tex]  

[tex]\theta=arctan(0.18)[/tex]  

[tex]\theta=10.20^{\circ}[/tex]  

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

[tex]P_{A}sin(\theta)-F_{fA}-F_{spring}=0[/tex]

Where:

[tex]F_{spring} = k\Delta l=2.1\Delta l[/tex]

[tex]P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0[/tex]

[tex]\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}[/tex]

[tex]\Delta l=0.10 ft[/tex]    

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

(a) The inclined angle for which both blocks begin to slide is 10.3⁰.

(b) The compression of the spring is 0.22 ft.

The given parameters;

The normal force on block A and B:

[tex]F_n_A = m_Agcos \ \theta\\\\F_n_B = m_Bgcos \ \theta[/tex]

The frictional force on block A and B:

[tex]F_f_A = \mu_s_AF_n_A \\\\F_f_B = \mu_s_BF_n_A[/tex]

The net force on the blocks when they starts sliding;

[tex](m_Ag sin \theta+ m_Bgsin\theta) - (F_f_A + F_f_B) = 0\\\\m_Ag sin \theta+ m_Bgsin\theta = F_f_A + F_f_B\\\\m_Ag sin \theta+ m_Bgsin\theta = \mu_Am_Agcos\theta \ + \ \mu_Bm_Bgcos\theta\\\\gsin\theta(m_A + m_B) = gcos\theta (\mu_Am_A + \mu_Bm_B)\\\\\frac{sin\theta}{cos \theta} = \frac{\mu_Am_A\ + \ \mu_Bm_B}{m_A\ + \ m_B} \\\\tan\theta = \frac{(0.16\times 11) \ + \ (0.23 \times 5)}{11 + 5} \\\\tan\theta = 0.1819\\\\\theta = tan^{-1}(0.1819)\\\\\theta = 10.3 \ ^0[/tex]

The change in the energy of the blocks is the work done in compressing the spring;

[tex]\Delta E = W\\\\F_A (sin \theta )d- \mu F_n d= \frac{1}{2} kd^2\\\\F_A sin\theta \ - \ \mu F_A cos\theta = \frac{1}{2} kd\\\\d = \frac{2F_A(sin\theta - \mu cos \theta) }{k} \\\\d = \frac{2\times 11(sin \ 10.3\ - \ 0.16\times cos \ 10.3) }{2.1} \\\\d = 0.22 \ ft[/tex]

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Answer:

1.The temperature of each sample will increase by the same amount

Explanation:

This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.

This is shown by the calculation below

Q = mcΔT

ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.

Since Q, m and c are the same for both substances, thus ΔT will be the same.

So, the temperature of each sample will increase by the same amount

Answer:

A) it transforms a small force acting over a large distance into a large force acting over a small distance.

Explanation:

The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.

Pressure transmitted P = F/A

where F is the force applied

and A is the area over which the force is applied.

This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.

F/A = f/a

where the left side of the equation is for the output, and the right side is for the input.

The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that

volume V = (area A) x (distance d)

this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.

From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.

Answer:

y ’= y / 2

thus when the slit width is doubled the pattern width is halved

Explanation:

The diffraction of a slit is given by the expressions

          a sin θ = m λ

where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.

          sin θ = m λ / a

If this equation

          a ’= 2 a

we substitute

          2 a sin θ'= m λ

          sin θ'= (m λ / a)  1/2

          sin θ ’= sin θ / 2

We can use trigonometry to find the width

         tan θ = y / L

as the angle is small

         tan θ = sin θ / cos θ = sin θ

         sin θ = y / L  

we substitute

        y ’/ L = y/L   1/2

        y ’= y / 2

thus when the slit width is doubled the pattern width is halved

Answer:

A. 990v/m

B.330x10^-8T

C.2.19x10^-6J/m³

D.1.45x10^-11J

Explanation:

See attached file

Answer:

A) 0.0267 T

B) 0.0263 T

Explanation:

Given that

The number of turns, N = 400

Radius of the wire, r = 6 cm = 0.06 m

Current in the wire, I = 0.25 A

Relative permeability, K(m) = 80

See the attached picture for the calculation

Answer:

-z axis

Explanation:

According to the left hand rule for an electron in a magnetic field, hold the thumb of the left hand at a right angle to the rest of the fingers, and the rest of the fingers parallel to one another. If the thumb represents the motion of the electron, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the electron. In this case, the left hand will be held out with the thumb pointing to the right (+x axis), and the palm facing your body (-y axis). The magnetic field indicated by the other fingers will point down in the the -z axis.

Answer:

E. None of these statements are true.

Explanation:

We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.

for A, the tension on the rope does not depend on if both teams pull are in equilibrium.

for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.

for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.

for D,  just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.

we go with option E.

Answer:

The wavelength is  [tex]\lambda = 419 \ nm[/tex]

Explanation:

From the question we are told that

   The  distance of separation is   [tex]d = 3.34 *10^{-5} \ m[/tex]

   The  distance of the screen is  [tex]D = 3.30 \ m[/tex]

      The  order of the fringe is  n =  7

     The distance of separation of  fringes is y =  29.0 cm = 0.29 m

Generally the wavelength of the light is mathematically represented as

          [tex]\lambda = \frac{y * d }{ n * D}[/tex]

substituting values

         [tex]\lambda = \frac{0.29 * 3.34*10^{-5} }{ 7 * 3.30}[/tex]

        [tex]\lambda = 4.19*10^{-7}\ m[/tex]

        [tex]\lambda = 419 \ nm[/tex]

Answer:

Explanation:

According to Equations of Projectile motion :

[tex]Time\ of\ Flight = \frac{2vsin(x)}{g}[/tex]

vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec

(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec

[tex]Maximum Height = \frac{(vsinx)^{2} }{2g}[/tex]

(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m

[tex]Horizontal Range = vcosx * t[/tex]

(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m

Answer:

The AC generator has one unbroken slip ring

Explanation:

In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.

An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.

We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.

It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.

Answer:

The direction of net magnetic force on the circular section of the loop is in the positive y-axis

The net magnetic force on the circular section of the loop is 3.74 N

Explanation:

The magnetic field strength [tex]B[/tex] = 1.7 T

the current [tex]I[/tex] = 0.7 A

The diameter of the loop = 2 m

the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2

==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m

The force on the semi-circular is given as

F = [tex]BIl[/tex] sin ∅

but the loop is perpendicular to the field, therefore

sin ∅ = sin 90° = 1

F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N

The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".

According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis

Answer:

[tex]q = -461532.5 \ C[/tex]

Explanation:

From the question we are told that

     The  electric filed is  [tex]E = 102 \ N/C[/tex]  

Generally according to Gauss law

=>   [tex]E A = \frac{q}{\epsilon_o }[/tex]

Given that  the electric field is pointing downward  , the equation become

    [tex]- E A = \frac{q}{\epsilon_o }[/tex]

Here   [tex]q[/tex] is the excess charge on the surface of the earth

          [tex]A[/tex] is the surface  area of the of the earth which is mathematically represented as

     [tex]A = 4\pi r^2[/tex]

Where r is the radius of the earth which has a value [tex]r = 6.3781*10^6 m[/tex]

 substituting values

    [tex]A = 4 * 3.142 * (6.3781*10^6 \ m)^2[/tex]

    [tex]A =5.1128 *10^{14} \ m^2[/tex]

So

   [tex]q = -E * A * \epsilon _o[/tex]

Here [tex]\epsilon_o[/tex] s the permitivity of free space with value

          [tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

     [tex]q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}[/tex]

     [tex]q = -461532.5 \ C[/tex]

Answer:

Explanation:

The formula for calculating the voltage across an inductor is expressed as

[tex]V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)[/tex]

Given parameters

current amplitude I = 1.50mA = 1.5*10⁻³A

inductance L = 0.450mH = 0.450*10⁻³H

Voltage across the inductor [tex]V_l[/tex] = 13.0V

Required

frequency f

Substituting the given parametres into the formula, we have;

[tex]V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz[/tex]

Hence, the frequency required is 3.067MHz

Answer:

Change in the frequency (in Hz) = 104.96 Hz

Explanation:

Given:

Speed of sound in air (v) = 343 m/s

Speed of car (v1) 36 m/s

Frequency(f) = 500 Hz

Find:

Change in the frequency (in Hz)

Computation:

Frequency hear by the observer(before)(f1) = [f(v+v1)] / v

Frequency hear by the observer(f1) = [500(343+36)] / 343

Frequency hear by the observer(f1) = 552.48 Hz

Frequency hear by the observer(after)(f2) = [f(v-v1)] / v

Frequency hear by the observer(f2) = [500(343-36)] / 343

Frequency hear by the observer(f2) = 447.52 Hz

Change in the frequency (in Hz) = f1 - f2

Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz

Change in the frequency (in Hz) = 104.96 Hz

Answer:

The current is  [tex]I = 8.9 *10^{-5} \ A[/tex]

Explanation:

From the question we are told that

     The  radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]

      The current density is  [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]

      The distance we are considering is  [tex]r = 0.5 R = 0.001585[/tex]

Generally current density is mathematically represented as

          [tex]J = \frac{I}{A }[/tex]

Where A is the cross-sectional area represented as

         [tex]A = \pi r^2[/tex]

=>      [tex]J = \frac{I}{\pi r^2 }[/tex]

=>    [tex]I = J * (\pi r^2 )[/tex]

Now the change in current per unit length is mathematically evaluated as

        [tex]dI = 2 J * \pi r dr[/tex]

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         [tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]

         [tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]

substituting values

        [tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]

        [tex]I = 8.9 *10^{-5} \ A[/tex]

Answer:

The speed of the proton is 4059.39 m/s

Explanation:

The centripetal force on the particle is given by;

[tex]F = \frac{mv^2}{r}[/tex]

The magnetic force on the particle is given by;

[tex]F = qvB[/tex]

The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.

[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]

where;

r is the radius of the circular path moved by both electron and proton;

⇒For electron;

[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]

⇒For proton

The speed of the proton is given by;

[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]

Therefore, the speed of the proton is 4059.39 m/s

Answer:

The charge is  [tex]Q = 2.177 *10^{-9} \ C[/tex]

Explanation:

From the question we are told that

     The electric potential is  [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]

     The  mass of the drop is  [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]

      The  speed is  [tex]v = 18.8 \ m/s[/tex]

Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question  is mathematically represented as

       [tex]Q = \frac{m v^2 }{ 2 * V }[/tex]

Substituting values

      [tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]

       [tex]Q = 2.177 *10^{-9} \ C[/tex]

Answer: no u

Explanation: no u

Explanation:

EE = ½ kx²

EE = ½ (800 N/m) (6 m)²

EE = 14,400 J

Answer:

14,400 J

Explanation:

Its the answer

Each electron winds up with kinetic energy of

(270 keV)

plus

(whatever KE it had when it started accelerating).

....................

Answer:

d. 107 cm above the water surface.

Explanation:

The refractive index of water and air = 1.333

The real height of the girl's sole above water = 80.0 cm

From the water, the apparent height of the girl's sole will be higher than it really is in reality by a factor that is the refractive index.

The boy in the water will therefore see her feet as being

80.0 cm x 1.333 = 106.64 cm above the water

That is approximately 107 cm above the water

Answer:use the formular distance over time i.e distance/time. Make sure to convert the distance from metres to kilometers and time from minutes to hours .

Explanation:

The average speed of the car is 31,680 meters per hour.

To calculate the average speed of the car, you need to divide the total distance traveled by the time it took to travel that distance.

Given:

Time taken (t) = 10 minutes = 10 minutes × (1 hour / 60 minutes) = 10/60 hours = 1/6 hours

Distance traveled (d) = 5,280 meters

Average Speed (v) = Distance (d) / Time (t)

Average Speed (v) = 5280 meters / (1/6) hours

To simplify, when you divide by a fraction, it's equivalent to multiplying by its reciprocal:

Average Speed (v) = 5280 meters × (6/1) hours

Average Speed (v) = 31,680 meters per hour

Hence, the average speed of the car is 31,680 meters per hour.

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Answer:

The blade undergoes 40 revolutions, so neither of the given options is correct!

Explanation:

The revolutions can be found using the following equation:

[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

Where:

α is the angular acceleration

t is the time = 2.5 s

[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min                

First, we need to find the angular acceleration:

[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]

Now, the revolutions that the blade undergo are:

[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]        

Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!

I hope it helps you!                              

Answer:

A) the moment of inertia of the system decreases and the angular speed increases.

Explanation:

The complete question is

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the  merry-go-round. As he does this, It is true to say that

A) the moment of inertia of the system decreases and the angular speed increases.

B) the moment of inertia of the system decreases and the angular speed decreases.

C) the moment of inertia of the system decreases and the angular speed remains the same.

D) the moment of inertia of the system increases and the angular speed increases.

E) the moment of inertia of the system increases and the angular speed decreases

In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as

[tex]I_{1} w_{1} = I_{2} w_{2}[/tex]    ....1

where [tex]I_{1}[/tex] and [tex]I_{2}[/tex] are the initial and final moment of inertia respectively.

and [tex]w_{1}[/tex] and [tex]w_{2}[/tex] are the initial and final angular speed respectively.

Also, we know that the moment of inertia of a rotating body is given as

[tex]I = mr^{2}[/tex]    ....2

where [tex]m[/tex] is the mass of the rotating body,

and [tex]r[/tex] is the radius of the rotating body from its center.

We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.

From equation 1, we see that in order for the angular momentum to be conserved, the decrease from [tex]I_{1}[/tex] to [tex]I_{2}[/tex] will cause the angular speed of the system to increase from [tex]w_{1}[/tex] to [tex]w_{2}[/tex] .

From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.