Category Archives: Engineering Mathematics Blog

College Physics by Openstax Chapter 6 Problem 24

Centripetal Force of a Rotating Wind Turbine Blade

Problem:

Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.

Solution:

We are given the following values:

  • radius, r=100 mr=100\ \text{m}
  • angular velocity, ω=0.5 rev/sec×2π rad1 rev=3.1416 rad/sec\omega = 0.5\ \text{rev/sec}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} = 3.1416\ \text{rad/sec}
  • mass, m=4 kgm=4\ \text{kg}

Centripetal force FcF_c is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity vv and has magnitude Fc=macF_c = m a_c which can also be expressed as

Fc=mv2ror Fc=mrω2F_c = m \frac{v^2}{r} \quad \text{or} \quad \ F_c = mr \omega^2

For this particular problem, we are going to use the formula Fc=mrω2F_c = mr \omega^2. If we substitute the given values, we have

Fc=mrω2Fc=(4 kg)(100 m)(3.1416 rad/sec)2Fc=3947.8602 NFc=4×103 N  (Answer)\begin{align*} F_c & =mr \omega^2 \\ \\ F_c & = \left( 4\ \text{kg} \right)\left( 100\ \text{m} \right)\left( 3.1416\ \text{rad/sec} \right)^2 \\ \\ F_c & = 3947.8602\ \text{N} \\ \\ F_c & = 4 \times 10^3\ \text{N}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The centripetal force on the end of the wind turbine blade is approximately 4×103 N4 \times 10^3\ \text{N}.

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College Physics by Openstax Chapter 6 Problem 23

The centripetal force of a child riding a merry-go-round

Problem:

(a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center?

(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?

(c) Compare each force with her weight.

Solution:

Part A

We are given the following values: m=22.0 kgm=22.0\ \text{kg}, ω=40.0 rev/min\omega = 40.0\ \text{rev/min}, and r=1.25 mr=1.25\ \text{m}. We are asked to solve for the centripetal force, FcF_c.

Centripetal force FcF_c is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity vv and has magnitude Fc=macF_c=m a_c, which can also be expressed as Fc=mv2rF_c = m \frac{v^2}{r} or Fc=mrω2F_c = m r \omega ^2. Basing from the given values, we are going to solve the problem using the formula

Fc=mrω2F_c = m r \omega ^2

First, we need to convert the angular velocity ω\omega to rad/sec\text{rad/sec} for unit homogeneity.

40 rev/min×2π rad1 rev×1 min60 sec=4.1888 rad/sec40\ \text{rev/min} \times \frac{2\pi\ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} = 4.1888\ \text{rad/sec}

Now, we can substitute the given values into our formula.

Fc=mrω2Fc=(22.0 kg)(1.25 m)(4.1888 rad/sec)2Fc=482.5162 NFc=483 N  (Answer)\begin{align*} F_c & = m r \omega ^2 \\ \\ F_c & = \left( 22.0\ \text{kg}\right) \left( 1.25\ \text{m}\right) \left( 4.1888\ \text{rad/sec}\right)^2 \\ \\ F_c & = 482.5162\ \text{N} \\ \\ F_c & = 483\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Let us convert the angular velocity to radians per second.

3.00 rev/min×2π rad1 rev×1 min60 sec=0.3142 rad/sec3.00\ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}}=0.3142 \ \text{rad/sec}

Now, we can substitute the given values

Fc=mrω2Fc=(22.0 kg)(8.00 m)(0.3142 rad/sec)2Fc=17.3750 NFc=17.4 N  (Answer)\begin{align*} F_c & = m r \omega ^2 \\ \\ F_c & = \left( 22.0\ \text{kg}\right) \left( 8.00\ \text{m}\right) \left( 0.3142\ \text{rad/sec}\right)^2 \\ \\ F_c & = 17.3750\ \text{N} \\ \\ F_c & = 17.4\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

For the first centripetal force we solved in Part A,

Fcw=483 N(22 kg)(9.81 m/s2)=2.24\frac{F_c}{w} = \frac{483\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 2.24

The centripetal force is 2.24 times the weight of the child.

For the centripetal force we solved in Part B, we have

Fcw=17.4 N(22 kg)(9.81 m/s2)=0.0806\frac{F_c}{w} = \frac{17.4\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 0.0806

The centripetal force is only about 8% of the child’s weight.

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College Physics by Openstax Chapter 6 Problem 21

Centripetal acceleration of an amusement park ride shaped like a Viking ship

Problem:

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity. The speed at the bottom of the arc is 23.4 m/s.

(a) What is the centripetal acceleration at the bottom of the arc?

(b) Draw a free-body diagram of the forces acting on a rider at the bottom of the arc.

(c) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.

(d) Discuss whether the answer seems reasonable.

Solution:

Problem 6-20: The centripetal acceleration of the commercial jet’s tires, and the force of a determined bacterium in it

At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m.

(a) At how many rev/min are the tires rotating?

(b) What is the centripetal acceleration at the edge of the tire?

(c) With what force must a determined 1.00×10−15 kg bacterium cling to the rim?

(d) Take the ratio of this force to the bacterium’s weight.

Solution:

We are given the following quantities: linear speed, v=60.0 m/s v=60.0 \ \text{m/s}, radius is half the diameter, r=0.425 m r=0.425 \ \text{m}.

Part A

We can compute the angular velocity based on the given using the formula, ω=vr \displaystyle \omega = \frac{v}{r}.

ω=vrω=60.0 m/s0.425 mω=141.1765 rad/sec\begin{align*} \omega & = \frac{v}{r} \\ \\ \omega & = \frac{60.0 \ \text{m/s}}{0.425 \ \text{m}} \\ \\ \omega & = 141.1765 \ \text{rad/sec} \end{align*}

Now, we can convert this into the required unit of rev/min.

ω=141.1765 radsec×1 rev2π rad×60 sec1 minω=1348.1363 rev/minω=1.35×103 rev/min  (Answer)\begin{align*} \omega & = 141.1765\ \frac{\text{rad}}{\text{sec}} \times \frac{1\ \text{rev}}{2\pi\ \text{rad}} \times \frac{60\ \text{sec}}{1\ \text{min}} \\ \\ \omega & = 1348.1363 \ \text{rev/min} \\ \\ \omega & = 1.35 \times 10^{3} \ \text{rev/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The centripetal acceleration at the edge of the tire can be computed using the formula, ac=rω2 a_{c} = r \omega ^{2}.

ac=rω2ac=(0.425 m)(141.1765 rad/sec)2ac=8470.5918 m/s2ac=8.47×103 m/s2  (Answer)\begin{align*} a_{c} & = r \omega ^2 \\ \\ a_{c} & = \left( 0.425\ \text{m} \right) \left(141.1765\ \text{rad/sec} \right)^2 \\ \\ a_{c} & = 8470.5918 \ \text{m/s}^2 \\ \\ a_{c} & = 8.47 \times 10 ^{3} \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

From the second law of motion, the force is equal to the product of the mass and the acceleration. In this case, we are going to use the formula, Fc=mac F_c = m a_c . We are given the mass to be m=1.00×10−15 kgm=1.00 \times 10 ^{-15}\ \text{kg} , and the centripetal acceleration is solved in Part B.

Fc=macFc=(1×10−15 kg)(8470.5918 m/s2)Fc=8.4705918×10−12 kg m/s2Fc=8.47×10−12 N  (Answer)\begin{align*} F_c & = ma_c \\ \\ F_c & = \left( 1 \times 10^{-15}\ \text{kg}\right) \left(8470.5918 \ \text{m/s}^2\right) \\ \\ F_c & = 8.4705918 \times 10 ^{-12}\ \text{kg m/s}^2 \\ \\ F_c & = 8.47 \times 10^{-12} \ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The ratio of this force, Fc F_c to the weight of the bacterium is

Fcmg=8.4705819×10−12 N(1×10−15kg)(9.81 m/s2)Fcmg=863.4640Fcmg=863  (Answer)\begin{align*} \frac{F_c}{mg} & = \frac{8.4705819 \times 10 ^{-12}\ \text{N}}{\left( 1 \times 10^{-15} \text{kg} \right)\left(9.81 \ \text{m/s}^2 \right)} \\ \\ \frac{F_c}{mg} & = 863.4640 \\ \\ \frac{F_c}{mg} & = 863 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Problem 6-19: The angular velocity of an “artificial gravity”

A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of 9.80 m/s2 at the rim?

Solution:

We are given the following quantities:

radius=diameter2=200 m2=100 m\text{radius} = \frac{\text{diameter}}{2} = \frac{200\ \text{m}}{2} = 100 \ \text{m}
centripetal acceleration,ac=9.80 m/s2\text{centripetal acceleration}, a_c = 9.80 \ \text{m/s}^2

Centripetal acceleration is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. The formula for centripetal acceleration is

ac=rω2a_{c} = r \omega ^2

If we solve for the angular velocity in terms of the other quantities, we have

ω=acr\omega = \sqrt{\frac{a_c}{r}}

Substituting the given quantities,

ω=acrω=9.80 m/s2100 mω=0.313 rad/sec  (Answer)\begin{align*} \omega & = \sqrt{\frac{a_c}{r}} \\ \\ \omega & = \sqrt{\frac{9.80 \ \text{m/s}^2}{100\ \text{m}}} \\ \\ \omega & = 0.313 \ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Problem 6-18: The linear speed of an ultracentrifuge and Earth in its orbit

Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.

(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).

Solution:

Part A

We are given a linear speed of an ultracentrifuge of 0.50 km/s0.50\ \text{km/s}. We are asked to verify this value if we are given a radius of r=0.100 mr=0.100\ \text{m} and angular velocity of ω=50000 rev/min \omega = 50000 \ \text{rev/min}. We are going to use the formula

v=rωv = r \omega

Since we are given a linear speed in km/s\text{km/s}, we are going to convert the radius to km\text{km}, and the angular velocity to rad/sec\text{rad/sec}

r=0.100 m×1 km1000 m=0.0001 kmr=0.100\ \text{m} \times \frac{1\ \text{km}}{1000\ \text{m}} = 0.0001\ \text{km}
ω=50000 rev/min×2π rad1 rev×1 min60 sec=5235.9878 rad/sec\omega = 50000 \ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} =5235.9878\ \text{rad/sec}

Now, we can substitute these into the formula

v=rωv=(0.0001 km)(5235.9878 rad/sec)v=0.5236 km/s  (Answer)\begin{align*} v & = r \omega \\ \\ v & = \left( 0.0001 \ \text{km} \right)\left( 5235.9878 \ \text{rad/sec} \right) \\ \\ v & = 0.5236 \ \text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

This value is about 0.500 km/s.

Part B

From Table 6.2 of the book

ParentSatelliteAverage orbital radius r(km)Period T(y)r3 / T2 (km3 / y2)
SunEarth1.496×1081.496 \times 10^{8} 13.35×10243.35 \times 10^{24}

Using the same formulas we used in Part A, we can solve for the linear velocity of the Earth around the sun. The radius is

r=1.496×108 kmr=1.496 \times10^{8} \ \text{km}

The angular velocity is

ω=1 revyear×2π rad1 rev×1 year365.25 days×1 day24 hours×1 hour3600 secω=1.9910×10−7 rad/sec\begin{align*} \omega & = 1 \ \frac{\text{rev}}{\text{year}} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1 \ \text{year}}{365.25 \ \text{days}} \times \frac{1\ \text{day}}{24\ \text{hours}}\times \frac{1\ \text{hour}}{3600\ \text{sec}} \\ \\ \omega & = 1.9910 \times 10^{-7}\ \text{rad/sec} \end{align*}

The linear velocity is

v=rωv=(1.496×108 km)(1.9910×10−7) rad/secv=29.7854 km/s  (Answer)\begin{align*} v & = r \omega \\ \\ v & = \left( 1.496\times 10^{8}\ \text{km} \right)\left( 1.9910 \times 10 ^ {-7} \right) \ \text{rad/sec}\\ \\ v & = 29.7854\ \text{km/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The linear velocity is about 30 km/s.

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Problem 6-17: The acceleration due to gravity at the position of a satellite located above the Earth

What percentage of the acceleration at Earth’s surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth?

Solution:

The acceleration due to gravity of a body and the Earth is given by the formula

g=GMr2g= G \frac{M}{r^2}

where GG is the gravitational constant, MM is the mass of the Earth, and rr is the distance of the object to the center of the Earth. We know that the approximate radius of the Earth is r=6.3781×106 mr=6.3781 \times 10^6 \ \text{m} .

The percentage of the acceleration at 300 km above the Earth of the acceleration due to gravity at Earth’s surface is

(GMr2)2(GMr2)1×100%\displaystyle \frac{\left( \frac{GM}{r^2} \right)_2}{\left( \frac{GM}{r^2} \right)_1} \times 100\%

Note that the subscript 2 indicates the satellite located 300 km above the Earth, and the subscript 1 indicates the object at the Earth’s surface. Also, from the expression above, we can cancel GG and MM from the numerator and denominator because these are constants. So, we are down to

(1r2)2(1r2)1×100%=(r2)1(r2)2×100%\frac{\left( \frac{1}{r^2} \right)_2}{\left( \frac{1}{r^2} \right)_1} \times 100\% = \frac{\left( r^2 \right)_1}{\left( r^2 \right)_2} \times 100\%

Substituting the values, we have

(r2)1(r2)2×100%=(6.3781×106 m)2(6.3781×106 m+300×103 m)2×100%=91.2172%=91.2%  (Answer)\begin{align*} \frac{\left( r^2 \right)_1}{\left( r^2 \right)_2} \times 100\% & = \frac{\left( 6.3781 \times 10^6 \ \text{m} \right)^{2}}{\left( 6.3781 \times 10^6 \ \text{m}+300 \times 10^{3} \ \text{m} \right)^{2}} \times 100\% \\ \\ & = 91.2172\% \\ \\ & = 91.2\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The percentage of the acceleration at the Earth’s surface of the acceleration due to gravity at the position of a satellite located 300 km above the Earth is about 91.2%.

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Problem 6-16: Calculating the centripetal acceleration of an ice skater’s nose

Olympic ice skaters are able to spin at about 5.00 rev/s.

(a) What is their angular velocity in radians per second?

(b) What is the centripetal acceleration of the skater’s nose if it is 0.120 m from the axis of rotation?

(c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since—at about 9.00 rev/s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius?

(d) Comment on the magnitudes of the accelerations found. It is reputed that Button ruptured small blood vessels during his spins.

Solution:

We are given an angular velocity, ω=5 rev/sec\omega = 5 \ \text{rev/sec}

Part A

For this part, we are asked to convert the angular velocity to units of radians per second.

ω=5.00 revsec×2π rad1 revω=31.4159 rad/secω=31.4 rad/sec  (Answer)\begin{align*} \omega & = \frac{5.00\ \text{rev}}{\text{sec}}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \\ \\ \omega & = 31.4159 \ \text{rad/sec} \\ \\ \omega & = 31.4 \ \text{rad/sec}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

For this part, we are asked to solve for the centripetal acceleration. We are going to use the formula ac=rω2a_{c} = r \omega ^2 given r=0.120 m r=0.120\ \text{m} and ω=31.4159 rad/s\omega = 31.4159 \ \text{rad/s} .

ac=rω2ac=(0.120 m)(31.4159 rad/s)2ac=118.4350 m/s2ac=118 m/s2  (Answer)\begin{align*} a_{c} & = r \omega ^2 \\ \\ a_{c} & = \left( 0.120 \ \text{m} \right) \left( 31.4159 \ \text{rad/s} \right)^2 \\ \\ a_{c} & = 118.4350 \ \text{m/s}^2 \\ \\ a_{c} & = 118 \ \text{m/s}^2\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

For this part, we are going to directly solve the centripetal acceleration.

ac=rω2ac=(0.120 m)(9 revs×2π rad1 rev)2ac=383.7302 m/s2ac=384 m/s2  (Answer)\begin{align*} a_{c} & = r \omega ^2 \\ \\ a_{c} & = \left( 0.120 \ \text{m} \right)\left( \frac{9\ \text{rev}}{\text{s}} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}}\right)^2 \\ \\ a_{c} & = 383.7302 \ \text{m/s}^2 \\ \\ a_{c} & = 384 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That is quite a lot of acceleration in itself. The centripetal acceleration felt by Button’s nose was 39.2 times larger than the acceleration due to gravity! It is no wonder that he ruptured small blood vessels in his spins.

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Curing the Compulsive Gambler: Challenging Probability Problem

Mr. Brown always bets a dollar on the number 13 at roulette against the advice of Kind Friend. To help cure Mr. Brown of playing roulette, Kind Friend always bets Brown $20 at even money that Brown will be behind at the end of 36 plays. How is the cure working?

(Most American roulette wheels have 38 equally likely numbers. If the player’s number comes up, he is paid 35 times his stake and gets his original stake back; otherwise, he loses his stake)

Solution:

If Mr. Brown wins once in 36 turns, he is even with the casino. His probability of losing all 36 times is (3738)36≈0.383\displaystyle \left( \frac{37}{38} \right)^{36} \approx 0.383 . In a single turn, his expectation is

35(138)−1(3738)=−238 dollars35\left( \frac{1}{38} \right)-1\left( \frac{37}{38} \right) = - \frac{2}{38}\ \text{dollars}

and in 36 turns

−238(36)=−1.89 dollars-\frac{2}{38}\left( 36 \right) = -1.89 \ \text{dollars}

Against Kind Friend, Mr. Brown has an expectation of

+20(0.617)−20(0.383)≈+4.68 dollars+20\left( 0.617 \right)-20\left( 0.383 \right)\approx +4.68 \ \text{dollars}

And so, all told, Mr. Brown gains +4.68 – 1.89 = + 2.79 dollars per 36 trials; he is finally making money at roulette. Possibly Kind Friend will be cured first. Of course, when Brown loses all 36, he is out $56, which may jolt him a bit.

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Chuck-a-Luck: Challenging Probability Problem

Chuck-a-Luck is a gambling game often played at carnivals and gambling houses. A player may bet on anyone of the numbers 1, 2, 3, 4, 5, 6. Three dice are rolled. If the player’s number appears on one, two, or three of the dice, he receives respectively one, two, or three times his original stake plus his own money back; otherwise, he loses his stake. What is the player’s expected loss per unit stake? (Actually, the player may distribute stakes on several numbers, but each such stake can be regarded as a separate bet.)

Solution:

Let us compute the losses incurred (a) when the numbers on the three dice are different, (b) when exactly two are alike, and (c) when all three are alike. An easy attack is to suppose that you place a unit stake on each of the six numbers, thus betting six units in all. Suppose the roll produces three different numbers, say 1, 2, 3. Then the house takes the three unit stakes on the losing numbers 4, 5, 6 and pays off the three winning numbers 1, 2, 3. The house won nothing, and you won nothing. That result would be the same for any roll of three different numbers.

Next suppose the roll of the dice results in two of one number and one of a second, say 1, 1, 2. Then the house can use the stakes on numbers 3 and 4 to payoff the stake on number 1, and the stake on number 5 to payoff that on number 2. This leaves the stake on number 6 for the house. The house won one unit, you lost one unit, or per unit stake you lost 1/6.

Suppose the three dice roll the same number, for example, 1, 1, 1. Then the house can pay the triple odds from the stakes placed on 2, 3, 4 leaving those on 5 and 6 as house winnings. The loss per unit stake then is 2/6. Note that when a roll produces a multiple payoff the players are losing the most on the average.

To find the expected loss per unit stake in the whole game, we need to weight the three kinds of outcomes by their probabilities. If we regard the three dice as distinguishable –say red, green, and blue — there are 6×6×6=2166 \times 6 \times 6= 216 ways for them to fall.

In how many ways do we get three different numbers? If we take them in order, 6 possibilities for the red, then for each of these, 5 for the green since it must not match the red, and for each red-green pair, 4 ways for the blue since it must not match either of the others, we get 6×5×4=1206 \times 5 \times 4 = 120 ways.

For a moment skip the case where exactly two dice are alike and go on to three alike. There are just 6 ways because there are 6 ways for the red to fall and only 1 way for each of the others since they must match the red.

This means that there are 216−126=90216 - 126 = 90 ways for them to fall two alike and one different. Let us check that directly. There are three main patterns that give two alike: red-green alike, red-blue alike, or green-blue alike. Count the number of ways for one of these, say red-green alike, and then multiply by three. The red can be thrown 6 ways, then the green 1 way to match, and the blue 5 ways to fail to match, or 30 ways. All told then we have 3×30=903 \times 30 = 90 ways, checking the result we got by subtraction.

We get the expected loss by weighting each loss by its probability and summing as follows:

120216×0⏟none alike+90216×16⏟2 alike+6216×26⏟3 alike=17216≈0.079\underbrace{\frac{120}{216}\times 0}_\text{none alike} + \underbrace{\frac{90}{216}\times \frac{1}{6}}_\text{2 alike}+\underbrace{\frac{6}{216}\times \frac{2}{6}}_\text{3 alike} = \frac{17}{216} \approx 0.079

Thus, you lose about 8% per play. Considering that a play might take half a minute and that government bonds pay you less than 4% interest for a year, the attrition can be regarded as fierce.

This calculation is for regular dice. Sometimes a spinning wheel with a pointer is used with sets of three numbers painted in segments around the edge of the wheel. The sets do not correspond perfectly to the frequencies given by the dice. In such wheels I have observed that the multiple payoffs are more frequent than for the dice, and therefore the expected loss to the bettor greater.

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