Uniform Circular Motion
First, radian and angular speed will be discussed.
Radian, the SI unit of Angle:
One radian of angle is the central angle in any circle which opposite arc is equal to the radius of that circle. If a piece of string is cut equal to the radius R of a circle and then placed on the edge of that circle, as shown, the central angle corresponding (or opposite) to that arc is called one "radian."
If a string, cut to length
R, is laid down on the edge
Example 1: Naming 3.14rd as " π ", calculate angles 360°, 180°, 90°, 60°, 45°, and 30° in terms of π.
|Fraction:||1 full circle||1/2 circle||1/4 circle||1/6 circle||1/8 circle||1/12 circle|
|Radians:||2π = 6.28rd||π||π/2||π/3||π/4||π/6|
Arc Length-Central Angle Formula:
There is an easy formula that relates any central angle ( θ ) to its opposite arc ( s ) and the radius ( R ) of a circle. This formula is valid only if the central angle is measured or expressed in radians.
Example 2: Referring to the above figure, suppose angle θ is 148° and R = 1.25 in. Calculate the length of arc AB.
Solution: Using S = R θ, and converting degrees to radians, yields: S = (1.25 in.)(148°)( 3.14rd / 180° ) = 3.23 in.
Angular Speed ( ω ):
The symbol "ω" is the lower case of symbol ( Ω ) pronounced "omega." Angular speed ( ω ) is defined as the change in angle per unit of time. Mathematically, it may be written as
ω = Δθ / Δt . The preferred unit for angular speed is rd /s. The commercial unit is (rpm) or revolutions per minute.
Note that each revolution is 6.28 radians.
Example 3: A car tire spins at 240rpm. Calculate (a) its angular speed in rd /s, (b) the angle that any of its radii sweeps in 44 seconds, and (c) the arc-length that any point on its outer edge travels during this time knowing that R = 14 in. Make sure that you completely solve this problem on paper using horizontal fraction bars everywhere.
Solution: 240rpm is the angular speed, ω. All we need to do in part (a) is to convert it from rpm to rd/s.
(a) ω = 240 (rev / min) = 240 ( 6.28 rd / 60sec) = 25 rd/sec. Note: 1rev = 6.28 rd, and 1min = 60 sec.
(b) ω = Δθ / Δt or Δθ = ωΔt or Δθ = ( 25 rd/sec)(44 sec) or, Δθ = 1100 rd.
(c) s = Rθ or s = ( 14 in.)(1100 rd) = 15400 in. = 1283 ft = 0.24 miles. (Verify the conversions).
Linear Speed - Angular Speed Formula:
The same way S and θ are related, we have a formula that relates v to ω. The formula is: v = Rω .
Writing these two similar formulas together helps their recall. s = Rθ and v = Rω.
The derivation is easy. All you need to do is to divide both sides of s = Rθ by (t) to get v = Rω.
(S/t) = R (θ /t) (S/t) is v, distance over time. (θ /t ) is ω, angle swept over time; therefore, v = Rω.
Example 4: The radius of a car tire is 14 in. Calculate (a) the linear speed (v) of any point on its outer edge if it spins at an angular speed of 25 rd/sec. (b) Find the linear distance (arc length) that any such point travels in 44 sec..
Solution: (a) Using v = Rω , yields: V = (14 in.)(25 rd/sec) = 350 in./sec.
(b) s = vt or s = (350 in./sec.)(44 sec.) = 15400 in. Use horizontal fraction bars (instead of slashes) when you solve.
Note: In (b) equation x = (1/2)at2 + vi t is used where a = 0 and x is replaced by s. This is because the arc length on the tire is like a long string wrapped around it and as the car moves, it leaves the unwrapped string on the ground as a straight line for which equation x = V t or s = V t is valid.
Chapter 5 Test Yourself 1:
1) Radian is a unit of (a) length (b) angle (c) area. click here
2) A central angle is an angle that has its vertex at the center of a (a) triangle (b) square (c) circle.
3) In any circle, the size of a central angle is equal to (a) the arc opposite to it when expresses in radians (b) half of the arc opposite to it (c) the radius of that circle.
4) Draw a circle and select two central angles in it, one equal to 90° and one equal to 45°. Since 90° is 1/4 of 360°, verify that the arc-length opposite to the 90°-angle you chose is also 1/4 of the whole circle. Also, verify that the 45°-angle you chose is opposite to an arc that is exactly 1/8 of the whole circle. What conclusion do you draw? State ......................Is your conclusion in line with the correct answer to Question 3? click here
5) 1rd is the central angle which opposite arc-length equals (a) the radius of the circle (b) the diagonal of the circle (c) the perimeter of the circle.
6) The central angle which opposite arc equals the perimeter of the circle is (a) 360° (b) 2π radians (c) both a and b.
7) The central angle which opposite arc equals half of the circle is (a) 120° (b) 180° (c) 3.14radians (d) both b and c.
8) 2.00 radians of angle is equivalent to (a) 114.6° (b) 120° (c) 170°. click here
9) S = Rθ calculates the arc length that is opposite to central angle θ only if θ is expressed in (a) degrees (b) radians (c) grads.
10) The arc length opposite to a 2.00rd central angle in a circle of radius R = 6.00" is (a) 12.0" (b) 8.00" (c) 18.0".
11) The arc length opposite to a 114.6° central angle in a circle of radius R = 6.00" is (a) 18.0" (b) 28.0" (c) 12.0".
12) Angular speed is defined as (a) the arc length traveled per unit of time (b) the angle traveled per unit of time (c) the angle swept by by a radius per unit of time (d) both b and c. click here
13) If angular change is Δθ and time change is Δt, then angular speed, ω, is (a) Δθ /Δt (b) Δt /Δθ (c) ΔθΔt.
14) The angular speed formula ω = Δθ /Δt is the counterpart of the linear speed formula v = Δx /Δt. (True) or (False)?
15) A spoke on a bicycle wheel travels or sweeps 150rd of angle every minute. Its angular speed is (a) 150rd/s (b) 150rd/min (c) 2.50rd/s (d) both b and c. click here
16) RPM is (a) a unit of angle (b) a unit of angular acceleration (c) a commercial unit of angular speed.
17) rd/s is (a) a commercial unit for angle (b) the SI unit for angular speed (c) neither a nor b click here
18) 180 rpm is the same thing as (a) 3.0 rps (b) 3.0 revolutions per second (c) 3.0 turns per second (d) a, b. and c.
19) 240 rpm is the same thing as (a) 240 turns per minute (b) 4.0 turns per second (c) both a and b.
20) 1 rpm is the same thing as 6.28 rd/min. True or False? click here
21) 3600 rpm is equivalent to (a) 60 rev/sec (b) 60x6.28rd / sec (c) 377 rd/s (d) a, b, and c.
22) A helicopter propeller rotates at 956 rpm. Its angular speed is (a) 100. rd/s (b) 200. rd/s (c) 300 rd/s. click here
23) The formula that relates linear speed to angular speed is (a) s = Rθ (b) v = Rω (c) x = vt + 1/2 at2.
24) A helicopter propeller rotates at 956 rpm. The linear speed of the tip of its propeller that is 5.00m long is (a) 500.m/s (b) 750m/s (c) 956m/s. click here
25) A helicopter propeller rotates at 956 rpm. The linear speed of the midpoint of its propeller that is 2.50m from its axis of rotation is (a) 500.m/s (b) 750m/s (c) 250m/s.
26) A helicopter propeller rotates at 956 rpm. The angular speed of the midpoint of it is (a) the same as the angular speed of the tip of it (b) is 1/2 of the angular speed of the tip of it (c) neither a nor b. click here
27) All points on a rotating solid wheel regardless of their radius of rotation have the same (a) linear speed (b) acceleration (c) angular speed.
28) The linear speed of a point on a solid rotating disk (a) is a function of R, its distance from the center of rotation (b) does not depend on R, its distance from the center (c) is a function of the angular speed of the disk. (d) both a and c.
29) All points on the outer edge of a rotating solid wheel have (a) the same angular speed (b) the same linear speed (c) the same linear velocity (d) a and b. click here
30) A solid wheel of radius 0.400m spins at 478 rpm. Find its angular speed in rd/s as well as the linear speed of points on it that are at radii 0.100m, 0.200m, and 0.300m.
Uniform Circular Motion
When a particle of mass M moves along a circular path at a constant rpm (revolutions per minute), its motion is called the "uniform circular motion."
Period of Rotation (T): The time it takes for the particle (mass M) to make one full turn is called the "period of rotation." T is the same thing as "seconds per turn."
Frequency (f): The number of turns per unit of time (usually per second) is called the frequency of rotation. (f) is the same thing as "turns per second."
From the definitions of T and f, it is easy to understand that the two are reciprocals.
T = 1/f or, f = 1/T
Example 5: A wheel is turning at 2 turns per second. Calculate the time for each turn or simply its period.
Solution: It is clear that each turn is made in (1/2)s; In other words, T = (1/2)s. This is the same thing as using the formula T=1/f and replacing (f) by 2 per second. If the frequency (f) is 3 per second, the time of each turn or (T) is T = (1/3)s.
Example 6: A car tire is spinning at a constant rate of 660rpm when the car's linear speed is 22m/s along the road. Determine (a) the frequency of rotation of the tire, (b) the period of rotation, and (c) the radius of the tire. Note: π = 3.14.
From this example, we may come up with a useful formula and that is
where V is the linear speed in circular motion.
Example 7: The Earth goes around the Sun once per year (as we know). Calculate the linear speed of the Earth as it goes around the Sun. The average distance between the two is 150 million kilometers. Suppose the orbit is a perfect circle although it is not.
Solution: The orbit is so big that its almost-circular-path feels like moving along a straight line. The radius of rotation, R, is given, but in (km). The period (T) is 1 year. Using the above formula:
V = ( 2πR) / T = ( 2π)(150,000,000km) / (365 x 24 x 3600s) = 30 km/s = 19 mi/s
The Relation Between ( f ) and ( ω ):
There is a useful formula that relates ω to f. Since f is the # of turns per second and ω the # of radians per second, and we know that ( 1turn = 2π radians); therefore, ω = 2π f .
Example 8: Solve the previous problem using ω = 2π f and v = Rω.
Solution: ω = 2π f or ω = 2π (1turn / year) or ω = 2π /year.
v = Rω or v = ( 150,000,000km) ( 2π /year ) = 30 km/s.
Chapter 5 Test Yourself 2:
1) In uniform circular motion, (a) the angular speed is constant (b) the linear speed is constant (c) the linear velocity is constant (d) the rpm is constant (e) a, b, and d. click here
2) In uniform circular motion, the radius of rotation can be varying and the motion still be uniform. (a) True (b) False
3) In any curved motion, including uniform circular motion, the velocity vector is always tangent to the path of motion at any point. (a) True (b) False click here
4) In uniform circular motion, the velocity vector at any point on the path is perpendicular to the radius of rotation at that point. (a) True (b) False.
5) The period of rotation is (a) the time for 1 radian of angle traveled (b) time for completing 100 turns (c) time for completing one turn. click here
6) Frequency, f, is (a) the # of turns per second (b) the # of rotations per second (c) the reciprocal of period, T (d) a, b, and c.
7) When f = 5s-1, it means that the spinning object makes 5 turns (a) per minute (b) per second (c) neither a, nor b.
8) When f = 5s-1, it is the same thing as writing (a) f = 5 /s (b) f = 5 rev/sec (c) f = 5 Hertz (d) f = 5 Hz (e) all of a, b, c, and d.
9) f = 10 rev/s is the same thing as (a) f = 98.6 rd/s (b) 31.4 rd/s (c) f = 10s-1 as well as ω = 62.8 rd/s. click here
10) f is the rotations frequency, but ω is the angular frequency that means the # of radians per second. (a) True (b) False
11) The formula that relates frequency f (the # of rotations per second) to ω (the # of radians swept per second) is ω=2πf. (a) True (b) False click here
12) A bicycle wheel makes 10 rotations per second.
We may write (a) f = 10s-1 and ω = 62.8 /s (b) f = 10s-1 and ω = 62.8 rd/s (c) f = 10s and ω = 62.8 rd/s.
13) A car tire runs at 480 rpm. We may write: click here
(a) f = 8s-1 and ω = 480 /s (b) f = 480s-1 and ω = 8 rd/s (c) f = 8/s and ω = 50.24 rd/s.
14) The angular speed, ω of a juicer is 314 rd/s. Its frequency, f is (a) 50.0/s (b) 50.0s-1 (c) 50.0Hz (d) a, b, & c.
15) A cul-de-sac has a radius of 60.0 ft. Two straight lines from the center of the cul-de-sac are drawn to its circular edge. The angle in between measures 36º. The arc length opposite to this central angle is (a) 1 rd (b) 3.6rd (c) 37.7ft.
16) The radius of a circle is 2.5m. The arc length opposite to 1rd of central angle in this circle is (a) 3.14m (b) 2.5m (c) 53.7º. click here
17) A driveway in a cul-de-sac of radius 60.0 ft shares 15ft of edge with that cul-de-sac. A kid draws two lines from the center of the cul-de-sac to the ends of that edge. The angle between these two lines is (a) 0.25 radians (b) 15π (c) 0.25ft.
18) The formula that relates arc-length S to central angle θ is (a) S = θ/R (b) S = Rθ (c) S = θ/2.
19) The formula that relates linear speed v to angular speed ω is v = Rω. (a) True (b) False click here
A piece of paper is stuck to the outer edge of a bicycle wheel that spins at 180 rpm. The radius of the wheel is 1.00ft.
20) The frequency of the rotation of the paper piece is (a) 18Hz (b) 3Hz (c) 0.5Hz.
21) The angular speed of the paper piece is (a) 25.14 rd/s (b) 18.84 rd/s (c) 18Hz. click here
22) The linear speed of the paper piece as it travels along the edge of the wheel is (a) 18.84 ft/s (b) 25.14 ft/s (c) 36 ft/min.
23) The linear distance that the paper piece travels in one second is (a) 3 circumferences (b) 6 circumferences (c) 0.
24) Each circumference is (a) πR2 (b) πR (c) 2πR.
25) Since the linear speed of the paper piece is 3 circumferences per second, it can be written as V = 3( 2πR ) /s. This is equal to writing V=R 2π(3/s) or, V = R2πf, or V=Rω. (a) True (b) False
(26) Since f = 1/T, the formula V=R2πf in Question 25 can be written as V = 2πR/T. (a) True (b) False
Speed in circular motion can be kept constant. Velocity is never constant in circular motion, because its direction keeps changing. Speed is constant only for uniform circular motion of a particle. As an example, when you turn a ceiling fan on, for a while, it is speeding up and therefore accelerating. During the acceleration phase, the motion is NOT uniform. When the fan reaches its maximum rpm, then each particle (or point) of it will have its own constant speed along its own circle of rotation.
In circular motion, there are two types of acceleration : "tangential acceleration" and "centripetal acceleration."
Tangential acceleration ( at ) is the result of the change in the speed of the object. In uniform circular motion where speed is constant, the tangential acceleration is zero.
Centripetal acceleration (ac ) is the result of the change in the direction of velocity of the object. This acceleration is never zero, because the change in direction is an ongoing process, in circular motion. The following figure shows two wheels: one spinning at constant rpm, and one at an increasing rpm. For the left wheel at = 0, and for the right wheel, at ≠ 0. However, for both wheels ac ≠ 0. Whether the wheel is turning at constant rpm or not, the centripetal acceleration is never zero.
The formula for centripetal acceleration is ac = v2/R . As was mentioned above, ac is a result of change in the direction of velocity vector with time. At this point, let's just accept this formula and solve a numerical example on it. We will then introduce "centripetal force," and also solve a couple of numerical examples on it, and finally investigate the reason why centripetal force as well as centripetal acceleration are directed toward the center of rotation.
Example 9: A tiny rock is caught in the treads of a car tire spinning at 420rpm on a wheel balancing machine. The radius of the tire is 0.35m. Calculate the centripetal acceleration given to the rock in forcing it to travel circularly.
Solution: ac = v2 / R. (v) must be calculated first. v = Rω ; ω = 2π f ; (f) means rev per second.
f = 420 (rev / min) = 420(rev / 60s) = 7.0 rev/s.
ω = 2π f ; ω = 2π ( 7.0 rev / s) = 44 rd /s.
v = Rω ; v = (0.35m)(44 rd/s) = 15.4 m/s.
ac = v2 / R ; ac = [15.4(m/s)]2 / 0.35m = 680 m/s2.
Example 10: A car has a centripetal acceleration of 2.3 m/s2 as it travels along a curved portion of a road. The road has a radius of curvature of 250m. Knowing that the linear speed of the car is constant, determine (a) its linear speed, (b) its angular speed, (c) the angle it travels in 6.0s, and (d) the fraction of the circle it travels during this period. The top view of the circular motion of the car is shown.
(a) ac = v2/R ; acR = v2 ; v =(acR)1/2 ; v = [2.3(m/s2) x 250m]1/2 = 24m/s.
(b) v = Rω ; v/R = ω ; ω = 24(m/s)/250m ; ω = 0.096rd/s (rd is borrowed)
(c) ω = Δθ/Δt ; ωΔt = Δθ ; Δθ = (0.096 rd/s)(6.0s) = 0.576rd
(d) Each circle is 6.28rd ; thus, fraction = 0.576rd /6.28rd = 0.092 = 9.2%
Centripetal Force: Objects tend to move along straight lines unless they are forced to do otherwise. When an object is moving along a circle, it is forced to have that circular path. The force pulls or pushes the object toward the center of rotation. That is why it is called the "centripetal force, Fc." When you tie a rock to a string and spin it above your head in a horizontal plane, your hand is constantly pulling the rock (via the string) toward your fingers that is the center of rotation. If you let go of the string, the centripetal force is eliminated and the rock will no longer follow that circular path; therefore, the cause of circular motion is the centripetal force, the force that pulls the object toward the center of rotation, no matter where the object is in its circular path. An object of mass M that has a centripetal acceleration, ac, along its circular path is therefore under a force of Fc = Mac. The formula for centripetal force becomes:
Fc = Mv2 / R
Example 11: A rock of mass 0.22kg is attached to a string of length 0.43m and is given a circular motion in a horizontal plane at a rate of 180rpm. Calculate the centripetal force that the string exerts on the rock pulling it nonstop toward the center of rotation. Draw a circle and show M, V, R , and Fc on it as shown above.
Solution: ω = 180 rev / min = 180( 6.28 rd / 60s) = 18.84 rd/s. (Not rounded)
v = Rω ; v = (0.43m)(18.84 rd/s) = 8.1 m/s.
Fc = Mv2 / R ; Fc = 0.22kg (8.1 m/s)2 / 0.43m = 34N.
A Good Link to Try: http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=21
Motion of a Car Along a Curved Road
Roads that are engineered (banked) provide a tilt (bank) in any of their curved portions that protects vehicles from slipping off the road specially under icy conditions. The tilt in a curved portion allows vehicles to safely have a greater speed along that portion compared to an un-tilted or un-banked portion of the same radius of curvature. We are going to derive a formula for the maximum speed of the motion of a vehicle along a curved portion for two cases: (1) for an un-banked road, and (2) for a banked road. The maximum speed is the speed at which the vehicle is on the verge of slipping off the road.
In case (1), the vehicle relies completely on friction, and in case (2) it relies completely on the road's tilt and not on friction at all. The real situation is a combination of both. Case (1) is studied in the following example and the formula is actually derived in there.
Example 12: An 840-kg car is negotiating a curve on an un-banked road where it has to totally rely on friction. The static coefficient of friction of the tires with the road is 0.73 and the radius of curvature is 110m. What maximum speed can the car have for not slipping off the road? In the first figure the car is being viewed from the back as it negotiates the road that curves to the left. In the second figure the top view is shown.
Motion of a Car Along an Un-banked Curved Road:
From the above example, we have come up with a general formula (*) for the motion of a car along an un-banked curved road, simply:
v2 = μs R g
As you may have noticed, the mass M of the car was given, but not used in the solution. It is interesting to note that the maximum speed does not depend on the mass of the vehicle, and for a fixed radius, it depends on μs only. Of course, no one should try to drive alongside a curve at maximum possible speed! There is no guarantee that the coefficient of static friction is the same everywhere along that curve.
Example 13: A car can negotiate an un-banked curve of radius 55m at a maximum speed of 66 km/h before the danger for slipping is felt. Determine the coefficient of friction that exists between its tires and the road.
Solution: v = 66 km/h ( 1000m / km )( 1h / 3600s ) = 18.3 m/s. (Use horizontal fraction bars).
v2 = μs R g ; μs = v2 / R g ; μs = (18.3m/s)2 / (55m X 9.8m/s2) = 0.62.
Motion of a Car Along a Banked Curved Road Without Relying on Friction:
When a vehicle travels along a banked curved road of radius ( R ), the tilt angle of the road is the important element that determines the maximum speed. Since the normal force, N , must be perpendicular to the road, it is not parallel to the weight force, w. In this case N > w. In fact, N adjusts itself such that its y-component neutralizes w and its x-component provides the needed centripetal force for the curved motion of the car to the left. The following is the force diagram from which maximum possible speed can be calculated:
Example 14: A car can travel along a banked curve of radius 125m at a maximum speed of 45 km/h without relying on friction. Determine the angle of the tilt of the road along this curve.
Solution: V = 45km/h = 45000m / 3600s = 12.5 m/s ; The appropriate formula is:
tan θ = V2 / Rg = 12.52 / (125x9.8) = 0.128 ; θ = 7.3º.
Chapter 5 Test Yourself 3:
1) In uniform circular motion, the linear speed is constant; therefore, the linear velocity is also constant. (a) True (b) False
2) In uniform circular motion, the velocity vector is always tangent to (a) the radius (b) the center (c) the circle.
3) The formulas s = Rθ and v = Rω are (a) both false (b) one true, one false (c) both true.
4) Tangential acceleration is (a) the change in angular speed per second (b) the change in the linear speed per second as an object speeds up along a circular path (c) the change in radius per second.
5) Centripetal acceleration is (a) due to changes in the direction of velocity vector that keeps changing in uniform circular motion (b) because of change in radius (c) none of a or b.
6) At any point on a circular path, the tangential and centripetal acceleration vectors are (a) parallel to each other (b) perpendicular to each other (c) normal to each other (d) b and c are the same and correct.
7) Centripetal acceleration vector that is always perpendicular to the path of motion and always directed toward the center of rotation, may also be called the normal acceleration. (a) True (b) False
8) Centripetal acceleration is proportional to (a) the magnitude of velocity, V (b) the square of velocity magnitude, V2 (c) the radius.
9) Centripetal acceleration is proportional to (a) the radius, R (b) the reciprocal of radius, 1/R (c) the radius squared, R2.
10) Overall, the formula that calculates the magnitude of the centripetal acceleration is (a) ac=V2R (b) ac=VR2 (c) ac=V2/R .
11) Since F = Ma ; therefore, (a) Fc = MV2/R (b) Fc = MV/R (c) Fc = MVR .
12) The velocity, V in the formula Fc = MV2/R can be found from the formula v = Rω. (a) True (b) False
13) It is reasonable to think that when driving along an un-banked curved road, the danger of slipping off the road is proportional to (a) the magnitude of velocity (b) the square of the magnitude of velocity (c) the reciprocal of velocity (1/V) because the faster we drive, the lesser the danger.
14) The formula that relates the coefficient of static friction to other factors for motion along an un-banked road where we completely rely on friction is (a) V = μs (b) V = μsRg (c) V2 = μsRg.
15) The reason for the use of μs, the coefficient of static friction, in the formula V2 = μsRg is that slipping should not occur. (a) True (b) False
16) For an engineered (banked) road, the greater the angle of tilt, (a) the lower the safe speed (b) the higher the safe speed (c) the smaller the coefficient of static friction.
17) For a banked curved road, the tilt angle is proportional to (a) the square of the magnitude of velocity (b) the magnitude of velocity (c) the reciprocal of velocity (1/V).
18) The tilt angle, θ of a banked road is (a) directly proportional to g (b) inversely proportional to g (c) is directly proportional to 1/g (d) both b & c.
19) The tilt angle, θ of a banked road is (a) directly proportional to R (b) inversely proportional to R (c) is directly proportional to 1/R (d) both b & c.
20) In the formula tan θ = V2 / Rg , (a) frictional effect is taken into account (b)no reliance is made on friction (c) if we include friction, it makes the angle of tilt greater.
1) A car can negotiate a banked portion of a road at a maximum speed of 25 mph under icy conditions. The radius of curvature of the road is 64m. Calculate the road's tilt angle.
2) The maximum speed that a car can make it through a curved portion of a flat (non-engineered) road is 46km/h. The radius of curvature of the road is 42m. Find the coefficient of static friction between the road and its tires.
3) A small plastic pail of water is attached to a string and spun in a vertical plane. What minimum speed is required at the top for the water not to spill out of the inverted pail? The radius of rotation of the water portion is 85cm.
4) A car travels along a road as shown. Both top and bottom of the hill have the same radius of curvature of 22m. (a) What is the maximum speed of the car for not to lose contact with the road at A? (b) If the car resumes the same maximum speed along the entire road, how heavy does an 83-kg person feel as the car travels at B? (c) What is the normal weight of the person in Newtons?
5) Calculate (a) the angular speed of the earth about its own axis that passes through its poles, and (b) the linear speed of the people who live on the equator. The radius of the earth is 6280km.
6) A metal cylinder has a radius of 25.0cm. At what minimum rpm must it spin about its vertical axis such that a small block of mass 40.0grams in contact with its vertical wall does not slide down? The coefficient of friction between the block and the wall of the cylinder is 0.650. Let g =9.81m/s2.
The condition for not falling of the block is Fs = w.
Answers: 1) 11° 2) 0.40 3) 2.9m/s 4) 33mph, 1620N, 810N 5) 7.27x10-5 rd/s, 456 m/s 6) 74.2 rpm