Uniform Circular Motion

Slide 1 / 112 Uniform Circular Motion 2009 by Goodman & Zavorotniy

Slide 2 / 112 Topics of Uniform Circular Motion (UCM) Kinematics of UCM Click on the topic to go to that section Period, Frequency, and Rotational Velocity Dynamics of UCM Vertical UCM Buckets of Water Rollercoasters Cars going over hills and through valleys Horizontal UCM Unbanked Curves Banked Curves Conical Pendulum

Slide 3 / 112 Key Terms and Equations Centripetal Force Equations: a R = v 2 /r F C = mv 2 /r Period, Frequency, & Rotational Velocity Equations: T = t/n = 1/f f = n/t = 1/T v = 2# r/t = 2# rf

Slide 4 / 112 Kinematics of UCM Return to Table of Contents

Slide 5 / 112 Kinematics of Uniform Circular Motion Uniform circular motion: motion in a circle of constant radius at constant speed Instantaneous velocity is always tangent to circle.

Slide 6 / 112 Kinematics of Uniform Circular Motion This acceleration is called the centripetal, or radial, acceleration, and it points towards the center of the circle.

Slide 7 / 112 Kinematics of Uniform Circular Motion Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that we get two similar triangles. A #l B A B r ## r r r C ## C

Slide 8 / 112 Kinematics of Uniform Circular Motion If displacement is equal to velocity multiplied by time, then vt is the displacement covered in time t. During that same time, velocity changed by an amount, Δv. r # vt v 1 r v 2 # v 1 v 2 #v

Slide 9 / 112 Kinematics of Uniform Circular Motion These are similar triangles because the angles are all congruent, so the sides must be in proportion. #v vt v = r #v v 2 = t r vt a = v 2 r # v 1 v 2 #v r r #

Slide 10 / 112 Kinematics of Uniform Circular Motion These are similar triangles because the angles are all congruent, so the sides must be in proportion. #v vt v = r Δv vt #v v 2 = t r a = v 2 r That's the magnitude of the acceleration. v 1 v 2 θ r θ r

Slide 11 / 112 Kinematics of Uniform Circular Motion Transpose v 2 to see the vector addition. v 1 v 1 v 2 v 2 #v θ θ The change in velocity, #v, shows the direction of the acceleration. In the picture, you can see #v points towards the center of the circle.

Slide 12 / 112 Kinematics of Uniform Circular Motion This acceleration is called the centripetal, or radial, acceleration. It direction is towards the center of the circle. It's magnitude is given by a = v 2 /r

Slide 13 / 112 1 Is it possible for an object moving with a constant speed to accelerate? Explain. A B C D No, if the speed is constant then the acceleration is equal to zero. No, an object can accelerate only if there is a net force acting on it. Yes, although the speed is constant, the direction of the velocity can be changing. Yes, if an object is moving it is experiencing acceleration. http://njc.tl/5y

Slide 14 / 112 2 Consider a particle moving with constant speed such that its acceleration of constant magnitude is always perpendicular to its velocity. A B C D It is moving in a straight line. It is moving in a circle. It is moving in a parabola. None of the above is definitely true all of the time. http://njc.tl/5z

Slide 15 / 112 3 An object moves in a circular path at a constant speed. Compare the direction of the object's velocity and acceleration vectors. A B C D Both vectors point in the same direction. The vectors point in opposite directions. The vectors are perpendicular. The question is meaningless, since the acceleration is zero. http://njc.tl/60

Slide 16 / 112 4 What type of acceleration does an object moving with constant speed in a circular path experience? A B C D free fall constant acceleration linear acceleration centripetal acceleration http://njc.tl/61

Slide 17 / 112 5 An object is traveling with a velocity of 6.0 m/s in a circular path whose radius is 4.0m. What is the magnitude of its centripetal acceleration? A 4 m/s 2 B 9 m/s 2 C 6 m/s 2 D 1.5 m/s 2 E.67 m/s 2 http://njc.tl/62

Slide 18 / 112 6 An object is traveling with a velocity of 6.0 m/s in a circular path. Its acceleration is 3.0 m/s 2. What is the radius of its path? A B C D E 6 m 1 m 2 m 16 m 12 m http://njc.tl/63

Slide 19 / 112 7 An object is traveling with a velocity in a circular path whose radius is 65 m. Its acceleration is 3.0 m/s 2. What is its velocity? A B C D E 11.87 m/s 15.36 m/s 19.74 m/s 13.96 m/s 195 m/s http://njc.tl/64

Slide 20 / 112 Period, Frequency, and Rotational Velocity Return to Table of Contents

Slide 21 / 112 Period The time it takes for an object to complete one trip around a circular path is called its Period. The symbol for Period is "T" Periods are measured in units of time; we will usually use seconds (s). Often we are given the time (t) it takes for an object to make a number of trips (n) around a circular path. In that case, T = t/n

Slide 22 / 112 8 If it takes 50 seconds for an object to travel around a circle 5 times, what is the period of its motion? A B C D E 5 s 10 s 15 s 20 s 25 s http://njc.tl/65

Slide 23 / 112 9 If an object is traveling in circular motion and its period is 7.0s, how long will it take it to make 8 complete revolutions? A B C D E 56 s 54 s 63 s 58 s 48 s http://njc.tl/66

Slide 24 / 112 Frequency The number of revolutions that an object completes in a given amount of time is called the frequency of its motion. The symbol for frequency is "f" Frequency are measured in units of revolutions per unit time; we will usually use 1/seconds (s -1 ). Another name for s -1 is Hertz (Hz). Frequency can also be measured in revolutions per minute (rpm), etc. Often we are given the time (t) it takes for an object to make a number of revolutions (n). In that case, f = n/t

Slide 25 / 112 10 An object travels around a circle 50 times in ten seconds, what is the frequency (in Hz) of its motion? A B C D E 25 Hz 20 Hz 15 Hz 10 Hz 5 Hz http://njc.tl/67

Slide 26 / 112 11 If an object is traveling in circular motion with a frequency of 7.0 Hz, how many revolutions will it make in 20 s? A 120 B 145 C 130 D 140 E 150 http://njc.tl/68

Slide 27 / 112 Period and Frequency Since T = t/n and f = n/t then T = 1/f and f = 1/T

Slide 28 / 112 12 An object has a period of 4.0s, what is the frequency of its motion (in Hertz)? A B C D E 1/16 Hz 1/8 Hz 1/4 Hz 1/2 Hz 2 Hz http://njc.tl/69

Slide 29 / 112 13 An object is revolving with a frequency of 8.0 Hz, what is its period (in seconds)? A B C D E 1/8 s 1/4 s 1/2 s 2 s 4 s http://njc.tl/6a

Slide 30 / 112 Rotational Velocity Each trip around a circle, an object travels a length equal to the circle's circumference. The circumference of a circle is given by: C = 2# r The time it takes to go around once is the period, T. And the object's speed is given by s = d/t So the speed must be: s = C/T = 2# r/t

Slide 31 / 112 Rotational Velocity A velocity must have a magnitude and a direction. The magnitude of an object's instantaneous velocity is its speed. So for an object in uniform circular motion, the magnitude of its velocity is: v = C/T = 2# r/t If an object is in uniform circular motion, the direction of its velocity is tangent to its circular motion. So v = 2# r/t tangent to the circle

Slide 32 / 112 14 An object is in circular motion. The radius of its motion is 2.0m and its period is 5.0s. What is its velocity? A B C D E 1.98 m/s 3. 57 m/s 4.36 m/s 3.25 m/s 2.51 m/s http://njc.tl/6b

Slide 33 / 112 15 An object is in circular motion. The radius of its motion is 2.0 m and its velocity is 20 m/s. What is its period? A B C D E 0.38 s 0.63 s 0.78 s 0.89 s 1.43 s http://njc.tl/6c

Slide 34 / 112 16 An object is in circular motion. The period of its motion is 2.0 s and its velocity is 20 m/s. What is the radius of its motion? A B C D E 7.87 m 3.56 m 5.61 m 6.36 m 5.67 m http://njc.tl/6d

Slide 35 / 112 Rotational Velocity Since f = 1/T, we can also determine the velocity of an object in uniform circular motion by the radius and frequency of its motion. v = 2# r/t and f = 1/T so v = 2# rf Of course the direction of its velocity is still tangent to its circular motion. So v = 2# rf tangent to the circle

Slide 36 / 112 17 An object is in circular motion. The radius of its motion is 2.0 m and its frequency is 8.0 Hz. What is its velocity? A B C D E 100.53 m/s 106.89 m/s 97.93 m/s 102.23 m/s 103.39 m/s http://njc.tl/6e

Slide 37 / 112 18 An object is in circular motion. The radius of its motion is 2.0 m and its velocity is 30 m/s. What is its frequency? A B C D E 4.12 Hz 2.82 Hz 2.39 Hz 3.67 Hz 1.78 Hz http://njc.tl/6f

Slide 38 / 112 19 An object is in circular motion. The frequency of its motion is 7.0 Hz and its velocity is 20 m/s. What is the radius of its motion? A B C D E 0.45 m 2.08 m 0.33 m 1.22 m 1.59 m http://njc.tl/6g

Slide 39 / 112 Dynamics of UCM Return to Table of Contents

Slide 40 / 112 Dynamics of Uniform Circular Motion For an object to be in uniform circular motion, there must be a net force acting on it. We already know the acceleration, so we can write the force:

Slide 41 / 112 Dynamics of Uniform Circular Motion We can see that the force must be inward by thinking about a ball on a string: Force on ball exerted by string Force on hand exerted by string

Slide 42 / 112 Dynamics of Uniform Circular Motion There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome. If the centripetal force vanishes, the object flies off tangent to the circle. This happens. This does NOT happen.

Slide 43 / 112 Curved Paths This concept can be used for an object moving along any curved path, as a small segment of the path will be approximately circular.

Slide 44 / 112 Centrifugation A centrifuge works by spinning very fast. This means there must be a very large centripetal force. The object at A would go in a straight line but for this force; as it is, it winds up at B.

Slide 45 / 112 20 What force is needed to make an object move in a circle? A B C D kinetic friction static friction centripetal force weight http://njc.tl/6h

Slide 46 / 112 21 When an object experiences uniform circular motion, the direction of the net force is A B C D in the same direction as the motion of the object in the opposite direction of the motion of the object is directed toward the center of the circular path is directed away from the center of the circular path http://njc.tl/6i

Slide 47 / 112 22 A car with a mass of 1800 kg goes around an 18 m radius turn at a speed of 35 m/s. What is the centripetal force on the car? http://njc.tl/6j

Slide 48 / 112 23 A 75 kg mass is attached to the end of a 5.0 m long metal rod string which rotates in a horizontal circular path. If the maximum force that the rod can withstand is 8500 N. What is the maximum speed that the mass can attain without breaking the rod? http://njc.tl/6k

Slide 49 / 112 Vertical UCM Return to Table of Contents

Slide 50 / 112 Car on a hilly road...

Slide 51 / 112 A car is traveling at a velocity of 20 m/s. The driver of the car has a mass of 60 kg. The car is located at the bottom of a dip in the road. The radius of the dip is 80m. What is the apparent weight of the driver (the normal force supplied by the seat of the car to support him) at the bottom of the dip?

Slide 52 / 112 This is a sketch of the problem. What do I do next? 1. Draw a free body diagram 2. Indicate the direction of acceleration v = 20 m/s m = 60 kg r = 80 m

Slide 53 / 112 a F N Next, mg 3. Draw axes with one axis parallel to acceleration v = 20 m/s m = 60 kg r = 80 m

Slide 54 / 112 y The dotted lines represent axes with one axis parallel to acceleration v = 20 m/s m = 60 kg r = 80 m a F N mg x All forces are parallel or perpendicular to the axes, so I don't have to resolve any vector into components

y Slide 55 / 112 4. Apply Newton's Second Law along each axis. x - direction y - direction a F N F = ma 0 = 0 F = ma F N - mg = ma F N = mg + ma F N = m (g + a) mg x While I don't know "a", I do know "v" and "r". What's my next step? v = 20 m/s m = 60 kg r = 80 m 5. Substitute a = v 2 /r F N = m (g + v 2 /r)

Slide 56 / 112 a F N 6. The last step. Substitute numbers. F N = m (g + v 2 /r) mg x F N = (60 kg) ((9.8 m/s 2 + (20 m/s) 2 / (80m)) F N = (60 kg) (9.8 m/s 2 + 5 m/s 2 ) v = 20 m/s m = 60 kg r = 80 m F N = (60 kg) (14.8 m/s 2 ) F N = (60 kg) (14.8 m/s 2 ) F N = 890 N How does this compare to his weight on the flat part of the road?

Slide 57 / 112 F N mg On the flat road the driver's weight (the normal force from the seat) is just mg. F N = mg F N = (60kg)(9.8 m/s 2 ) = 590 N Apparent Weight Flat road Bottom of dip (r = 80m) v = 20 m/s m = 60 kg r = 590 N 890 N The gravitational force on the driver (mg) doesn't change, but his apparent weight (F N ) does. Is there a situation where he will appear weightless?

Slide 58 / 112 At what velocity must a car drive over a hill if the driver (and the car for that matter) are to appear weightless? The driver of the car has a mass of 60 kg. The radius of the hill is 80m.

Slide 59 / 112 This is a sketch of the problem. What do I do next? 1. Draw a free body diagram 2. Indicate the direction of acceleration m = 60 kg r = 80 m

Slide 60 / 112 m = 60 kg r = 80 m This is the free body diagram (let's start out with F N, knowing we'll eventually let it equal zero) Next, a F N mg 3. Draw axes with one axis parallel to acceleration

m = 60 kg r = 80 m y Slide 61 / 112 The dotted lines represent axes with one axis parallel to acceleration F N a mg x

Slide 62 / 112 m = 60 kg r = 80 m y 4. Apply Newton's Second Law along each axis. x - direction y - direction a F N mg x ΣF = ma 0 = 0 #F = ma F N - mg = -ma F N = mg - ma F N = m (g - a) But I'd like to find the velocity that the car must be going for the driver (and car) to appear weightless. How? 5. Substitute a = v 2 /r F N = m (g - v 2 /r)

m = 60 kg r = 80 m y Slide 63 / 112 6. The last step, when does F N = 0 When the product of two variables is zero, then one of them must be zero. Since m is not zero, the only way F n can equal zero is... F N a mg x F N = m (g - v 2 /r) 0 = (60 kg) (g - v 2 /r) 0 = (g - v 2 /r) v 2 /r = g v = (gr) 1/2 v = ((9.8m/s 2 )(80m)) 1/2 v = 28 m/s

Slide 64 / 112 Practice Problem: A car is traveling at a velocity of 15 m/s. The driver of the car has a mass of 50 kg. 240 N The car is located at the top of a hill in the road. The radius of the hill is 45 m. What is the apparent weight of the driver (the normal force supplied by the seat of the car to support him) at the top of the hill?

Slide 65 / 112 24 A car is going over the top of a hill whose curvature approximates a circle of radius 175 m. At what velocity will the occupants of the car appear to weigh 10% less than their normal weight (F N = 0.9mg)? A B C D E 13.1 m/s 14. 7 m/s 13.9 m/s 14.2 m/s 12.7 m/s http://njc.tl/6l

Slide 66 / 112 25 A car is going through a dip in the road whose curvature approximates a circle of radius 175 m. At what velocity will the occupants of the car appear to weight 10% more than their normal weight? A B C D E 9.8 m/s 12.7 m/s 11.9 m/s 13.1 m/s 14.5 m/s http://njc.tl/6m

Slide 67 / 112 26 The occupants of a car traveling at a speed of 25 m/s note that on a particular part of a road their apparent weight is 20% higher than their weight when driving on a flat road. Is that part of the road a hill, or a dip? A B Hill Dip http://njc.tl/6n

Slide 68 / 112 27 The occupants of a car traveling at a speed of 25 m/s note that on a particular part of a road their apparent weight is 20% higher than their weight when driving on a flat road. What is the vertical curvature of the road? A B C D E 345.67 m 298.74 m 276.91 m 399.35 m 318.88 m http://njc.tl/6o

Slide 69 / 112 Buckets and Rollercoasters...

Slide 70 / 112 A bucket of water is being spun is a vertical circle of radius 0.80m. What is the smallest velocity which will result in the water not leaving the bucket? What do you know and what are you looking for? r = 0.80 m g = 9.8 m/s 2 downward v =?

Slide 71 / 112 r = 0.80 m g = 9.8 m/s 2 downward v minimum =?

Slide 72 / 112 mg a T r = 0.80 m g = 9.8 m/s 2 downward v minimum =?

Slide 73 / 112 mg ΣF = ma T r = 0.80 m g = 9.8 m/s 2 downward v minimum =?

Slide 74 / 112 mg ΣF = mv 2 /r T r = 0.80 m g = 9.8 m/s 2 downward v minimum =?

Slide 75 / 112 mg T ΣF = mv 2 /r ΣF = ma T + mg = m(v 2 /r) T = m(v 2 /r) - mg T = m(v 2 /r - g) T = 0 for minimum v v 2 /r - g = 0 v 2 /r = g v = (gr) 1/2 = ((9.8)(0.8)) 1/2 = 2.8 m/s r = 0.80 m g = 9.8 m/s 2 downward v minimum =?

Slide 76 / 112 Assuming a constant speed, and that the mass of the bucket is 2.5kg, what will the tension in the string be at the bottom of the circle? T r = 0.80 m g = 9.8 m/s 2 downward

Slide 77 / 112 Assuming a constant speed, and that the mass of the bucket is 2.5kg, what will the tension in the string be at the bottom of the circle? ΣF = ma T - mg = ma T T - mg = m(v 2 /r) ΣF = mv 2 /r T = m(v 2 /r) + mg T = m(v 2 /r + g) T = 2.5kg (9.8 m/s 2 + 9.8 m/s 2 ) mg T = 2.5kg (19.6 m/s 2 ) T = 490 N r = 0.80 m g = 9.8 m/s 2 downward

Slide 78 / 112 Practice Problem: A bucket of water is being spun is a vertical circle of radius 1.2 m. What is the smallest velocity which will result in the water not leaving the bucket? 3.43 m/s Assuming a constant speed, and that the mass of the bucket is 1.3 kg, what will the tension in the string be at the bottom of the circle? 25.48 N

Slide 79 / 112 28 A ball is attached to the end of a string. It is swung in a vertical circle of radius 10 m. What is the minimum velocity that the ball must have to make it around the circle? A B C D E 2.8 m/s 9.9 m/s 7.5 m/s 2.1 m/s 3.9 m/s http://njc.tl/6p

Slide 80 / 112 29 A ball is attached to the end of a string. It is swung in a vertical circle of radius 2.25 m. What is the minimum velocity that the ball must have to make it around the circle? A B C D E 3.87 m/s 4.34 m/s 4.69 m/s 5.12 m/s 5.39 m/s http://njc.tl/6q

Slide 81 / 112 30 A roller coaster car is on a track that forms a circular loop in the vertical plane. If the car is to just maintain contact with track at the top of the loop, what is the minimum value for its centripetal acceleration at this point? A B C D g downward 0.5g downward g upward 2g upward http://njc.tl/6r

Slide 82 / 112 31 A roller coaster car (mass = M) is on a track that forms a circular loop (radius = r) in the vertical plane. If the car is to just maintain contact with the track at the top of the loop, what is the minimum value for its speed at that point? A rg B (rg) 1/2 C (2rg) 1/2 D (0.5rg) 1/2 http://njc.tl/6s

Slide 83 / 112 32 A pilot executes a vertical dive then follows a semi-circular arc until it is going straight up. Just as the plane is at its lowest point, the force on him is A B C D less than mg and pointing up less than mg and pointing down more than mg and pointing up more than mg and pointing down http://njc.tl/6t

Slide 84 / 112 Horizontal UCM Return to Table of Contents

Slide 85 / 112 Banked and Unbanked Curves When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.

Slide 86 / 112 Banked and Unbanked Curves If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.

Slide 87 / 112 Banked and Unbanked Curves As long as the tires do not slip, the friction is static. If the tires do start to slip, the friction is kinetic, which is bad in two ways: The kinetic frictional force is smaller than the static. The static frictional force can point towards the center of the circle, but the kinetic frictional force opposes the direction of motion, making it very difficult to regain control of the car and continue around the curve.

Slide 88 / 112 Unbanked Curves A car is going around a track with a velocity of 20 m/s. The radius of the track is 150 m. What is the minimum coefficient of static friction that would make that possible? What do you know and what are you looking for? v = 20 m/s r = 150 m µ =?

Slide 89 / 112 Unbanked Curves Top View Front View (the car heading towards you) r v = 20 m/s r = 150 m µ =?

Slide 90 / 112 Unbanked Curves Top View Front View (the car heading towards you) F N r a f s v mg v = 20 m/s r = 150 m µ =?

Slide 91 / 112 Unbanked Curves vertical F N v a r mg f s radial v = 20 m/s r = 150 m µ =?

vertical Slide 92 / 112 Unbanked Curves vertical direction radial direction F N f s ΣF = ma F N mg = 0 F N = mg radial ΣF = ma f s = ma μ s F N = m(v 2 /r) μ s mg = mv 2 /r mg μ s = v 2 /gr μ s = (20m/s) 2 /((9.8 m/s2)(150m)) μ s = 0.27 v = 20 m/s r = 150 m µ =?

Slide 93 / 112 33 A car goes around a curve of radius r at a constant speed v. Then it goes around the same curve at half of the original speed. What is the centripetal force on the car as it goes around the curve for the second time, compared to the first time? A B C D twice as big four times as big half as big one-fourth as big http://njc.tl/6u

Slide 94 / 112 Banked Curves Banking curves can help keep cars from skidding. In fact, for every banked curve, there is one speed where the entire centripetal force is supplied by the horizontal component of the normal force, and no friction is required. Let's figure out what that speed is for a given angle and radius of curvature.

Slide 95 / 112 Banked Curves Top View Front View (the car heading towards you) y a r v a x We know the direction of our acceleration, so now we have to create axes. mg # Note that these axes will be different than for an Inclined Plane, because the car is not expected to slide down the plane. Instead, it must have a horizontal acceleration to go in a horizontal circle.

Slide 96 / 112 Banked Curves vertical Next, do the free body diagram. y radial a Let's assume that no friction is necessary at the speed we are solving for. x # mg

Slide 97 / 112 Banked Curves vertical Next, decompose the forces that don't align with an axis, F N. # F N y radial a x # mg mg

Slide 98 / 112 Banked Curves F N cos# vertical # F N F N sin# radial a Now let's solve for the velocity such that no friction is necessary to keep a car on the road while going around a curve of radius r and banking angle #. mg

Slide 99 / 112 Banked Curves vertical vertical direction ΣF = ma radial direction ΣF = ma F N cos# # F N F N sin# F N cos# mg = 0 F N = mg/cos# F N sin# = ma (mg/cos#)(sin#) = m(v 2 /r) (g/cos#)(sin#) = (v 2 /r) a radial (gtan#) = (v 2 /r) v = (grtan#) 1/2 mg

Slide 100 / 112 Practice Problem: Determine the velocity that a car should have while traveling around a frictionless curve with a radius 250 m is banked at an angle of 15. 25.6 m/s

Slide 101 / 112 Conical Pendulum A Conical Pendulum is a pendulum that sweeps out a circle, rather than just a back and forth path. Since the pendulum bob is moving in a horizontal circle, we can study it as another example of uniform circular motion. However, it will prove necessary to decompose the forces.

Slide 102 / 112 Conical Pendulum # Draw a sketch of the problem, unless one is provided. l Then draw a free body diagram and indicate the direction of the acceleration.

Slide 103 / 112 Conical Pendulum # l T a mg Then draw axes with one axis parallel to acceleration

Slide 104 / 112 Conical Pendulum # l T a mg Then decompose forces so that all components lie on an axis...in this case, decompose T.

Slide 105 / 112 Conical Pendulum Tcos# # Tsin# mg a Then solve for the acceleration of the bob, based on the angle #, by applying Newton's Second Law along each axis.

Slide 106 / 112 Tcos# Conical Pendulum x - direction F = ma Tsinθ = ma y - direction F = ma Tcosθ - mg = 0 Tcosθ = mg # Tsin# mg a Divide these two results Tsin θ = ma Tcos θ = mg tan θ = a/g θ = tan -1 (a/g) Or a = g tan θ

Slide 107 / 112 Conical Pendulum # mg T a An alternative approach is to solve this as a vector equation using #F=ma. (This will work whenever only two forces are present.) Just translate the original vectors (before decomposing them) to form a right triangle so that the sum of the two forces equals the new vector "ma".

Slide 108 / 112 Conical Pendulum Then, since tan # = opposite/adjacent # T ma mg tan # = ma/mg = a/g tan # = (v 2 /r)/g = v 2 /gr v 2 = grtan # the same result we found before

Slide 109 / 112 l # A 0.5 kg ball on a string is whirled in a circle with a radius of 0.75 m with a speed of 3 m/s. Find the tension in the string.

Slide 110 / 112 A 0.5 kg ball on a string is whirled in a circle with a radius of 0.75 m with a speed of 3 m/s. a Find the tension in the string. T T y x-direction y-direction T x #F = ma #F = ma Tx = ma Ty - mg = 0 Tx = mv 2 /r Ty = mg mg T 2 = Tx 2 + Ty 2 T 2 = (mv 2 /r) 2 + (mg) 2 T 2 = m 2 (v 4 /r 2 + g 2 ) T 2 = (0.5kg) 2 ((3m/s) 4 /(0.75m) 2 + (9.8m/s 2 ) 2 ) T 2 = 60 N 2 T = 7.7 N

Slide 111 / 112 l # A 1.5 kg ball on a string is whirled in a circle with a radius of 2.25 m with a speed of 6 m/s. Find the tension in the string. 28.14 N

Slide 112 / 112 Nonuniform Circular Motion If an object is moving in a circular path but at varying speeds, it must have a tangential component to its acceleration as well as the radial one.