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CIRCULAR MOTION
1. Motion in a circle -
a body which travels equal distances in equal times along a circular path has constant speed but not constant velocity.
-
since the direction of the velocity is changed from time to time, the body has an acceleration.
2. Angles in radians θ Consider a body moves in a circular path with centre 0. s
=
θ
r
s = distance travelled r = radius of the circle
= angular displacement 3. Angular velocity
w - it is the angle swept out in unit time by the radius joining the body to the centre of the circle. (Unit : rad s-1 ) - Consider a body moving uniformly from A to B in time t
w
t
=
s rt
=
v
v =
4.
=
T
=
T
=
r
( v = constant speed of the body )
rw 2 r / v 2 w
( T = period of the circular motion )
Centripetal acceleration in a circular motion -
consider a body moving with constant speed v in a circle of radius r and centre .
It
travels from A to B in a short interval of time t .
magnitude of v A and v B = v -
If t is very small, will also be small and CD will have have almost the same length as arc CD.
CIRCULAR MOTION MOTION | Gayan SK Perera
2 v
v
v a
t
v
t
vw
a = v2 /r = rw2 -
The direction of the acceleration is towards centre O of the circle as can be seen if t is made so small that A and B all but coincide; vector CD is then perpendicular to V A (or V B)
i.e. along AO (or BO). -
We say the body has centripetal acceleration.
This centripetal acceleration is not constant since its direction is varied from time to time.
5. Centripetal force -
We know from Newton ’s second law that whenever a body is accelerating, there must be a resultant force.
If a body of mass m follows a circular path of radius r with speed v,
its acceleration is v2/r and it must be subject to a centripetal force.
mv 2 F mrw 2 r -
The acceleration is directed towards the centre of circle so the force, too, must act in this direction.
e.g.1
For this reasons the force is called a centripetal force.
A ball of mass m attached to a string is swung round in a HORIZONTAL CIRCLE with radius r and uniform speed v.
T -
mv r
tension of the string T provides the necessary centripetal force so that the ball can be maintained in a circular path.
e.g.2
A mass is swung in a VERTICAL CIRCLE with uniform speed v and radius r.
i. .
0°<θ<90° T mg cos
mv 2 r
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T mg cos
(II)
mv 2 r
90°<θ<180° T mg cos(180 )
T mg cos T mg cos T ** N.B. (1)
For
mv 2 r
mv 2 r
mv 2 r mv 2 r
mg cos
90° < θ < 180° T = 0 when - mg cos
mv 2 r
i.e. v2 = - gr cos θ
maximum tension T mg
(2)
minimum tension T
e.g.3
mv r
mv r
,
when
θ=
0°
mg , when θ= 180°
i.e. at B i.e. at A
Overturning of a car
R 1 + R 2 = mg
………..(1)
take moment about c.g., (R 2 – R 1) a
= f h ………..(2)
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from (1), (2) R 2 = R 1 =
1 2 1 2
(mg (mg
fh a fh
) )
a
** Overturning just begin when R 1 = 0, R 1= 0
fh
i.e. mg since
…………………(3)
a
mv 2
f ………………(4)
r
sub. (4) into (3), mg v2
mv 2 h r a
gar h
(condition for overturning just begin)
Hence, the general condition for overturning is v2
Question:
gar h
How can we design a car so that it can have a large velocity but without overturning?
6. Other applications with centripetal forces a) The conical pendulum – The figure below shows a conical pendulum in which a mass m, supported by a string of length
,
follows a horizontal circular
path of radius r with constant speed v.
T cos θ = mg ………(1) T sin θ = ( 2) (1)
tan
mv 2 r
……….(2)
v2 rg
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** N.B. (I)
the horizontal component of the tension in string provides the required centripetal force.
(II)
tan
θ=
v 2 / rg
i.e. at a fixed angle θ, v2 is directly proprotional to the
radius of the circle r.
b)
Rounding a bend (i)
without banking - A car is travelling round a circular bend with uniform speed on a horizontal road. - Air resistance is always against the instantaneous motion of the car.
The
other and more important horizontal force is the frictional force exerted inwards by the ground on the tyres of the car.
-
The centripetal force, this case is due to the resultant interaction of the car with air and the ground.
*N.B. (i)
the circular motion can be maintained only if the frictional force can provided the required centripetal force -
otherwise skidding occurs,
radius of the circular motion will then to increased. (ii) safe cornering that does not role on friction is achieved by ‘ banking’ the road.
(ii)
With banking a car is moving round a banked track in a circular path of horizontal radius r. If the only forces at the wheels are the normal reaction N (no frictional force are introduced), then
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N co s N sin ( 2) (1)
**N.B. (i)
mg ....................(1) mv 2
tan
r
....................( 2)
v2 gr
the centripetal force is only due to the horizontal component of the normal reaction on the wheels by the road.
(ii)
In this case, a car can have speed v without skidding if the banked angle of the road is θ s.t.
tanθ = v 2/rg.
Question: What happens when (1) (2)
tanθ > v 2/rg
,
and
tanθ < v 2/rg ?
c.
Aircraft turning in flight
-
An aricraft in straight, level flight experiences a lifting force at right angles to the surface of its wings which balances its weight.
-
To turn, the aircraft is banked and (i)
the horizontal component of the lifting force supplies the necessary centripetal force.
(ii) Now only the vertical component of the lifting force support the aircraft ’s weight and height will be lost unless the lifting force if increased by, for example, increasing the speed.
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d. Looping the loop -
A bucket of water can be swung round in a vertical circle without spilling.
(i)at position A, mv 2 r
mg N
** if mg is greater than Hence,
mg
mv 2 r
mv 2 r
at A, the water will not stays in the bucket.
is required in this case.
(ii) at position B, mv 2 r
N
(iii) at position C, mv 2 r
N mg
(iv) at position D, mv2 r
Question:
N
For a fixed speed v and fixed radius r, Which position will have (1) (2)
greatest N, and smallest N
e. Centrifuges (i)
Introduction – they are used to separate solids suspended in liquids of different densities. - when the mixture in a tube is rotated at high speed in a horizontal
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circle, the loss dense matter moves towards the centre of rotation.
(ii) Explanation – P 1>P2 its that its difference can provide the necessary centripetal force acting radially inwards. - Thus PB>PA and there is a pressure gradient along the tube.
And
hence any part of the liquid experiences an inward force. Consider two particles X,Y in the liquid so that Mx < My. For a constant centripetal force and rotating speed. we have F F
Mxv2 r x Myv2 r y
........................(1)
........................(2)
since (1) = (2), -
Mx r x
My r y
(i.e. M r close to the centre of rotation)
Hence, suspended matter of smaller density will move inwards because such inward force is too large to it comparatively.
f.
Rotor (i)
Introduction – it consists of an upright drum.
The drum is spun at increasing
speed about its central vertical axis and at a certain speed the floor is pulled downwards.
The occupants do not fall but
remain ‘ pinned’ against the wall of the rotor.
N = normal reaction on the occupant by the wall f = frictional force acting on the occupant by the wall (ii) Explanation –
if the frictional force acting on the man by the will can
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balance his weight, then he will not fall down. N
mv 2
r f mg
...............(1)
N mg ...............(2) ( 2) (1)
**N.B.
(1)
gr v2
sufficient frictional force can be obtained if
gr v
2
i.e. min
gr v
2.
(2) the man’s weight has no effect to his result.
g. Motion of bicycle rider round circular track (i)
When a person on a bicycle rides round a circular racing racing track with radius r., the frictional force (provides the centripetal force)
f has a moment about the c.g.
which tends to turn the rider outwards.
(ii) When the rider leans inwards, moment of f is counter-balanced by the moment of N about c.g. M = mass of bicycle and rider f = frictional force acting on the wheel by the ground N = normal reaction on the wheel by the ground v = speed of the bicycle take moment about c.g., f h = N a ………….(1)
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resolve horizontally,
f
resolve verticall y,
N mg ..................(3) mv 2
sub. (2), (3) into (1),
r a h
(i)
r
..................(2)
h mg a v2 rg
tan
**N.B.
mv 2
v2 rg
θ =
angle of inclination to the vertical
at a circular track of fixed r,
tan v 2 (ii)
to avoid skidding, f
mv 2 r
mg
mv 2 r
v2 gr
tan
condition to avoid skidding.
Problems
1.
An object of mass 4 kg moves round a circle of radius 6 m with a constant speed of 12 m s_1 Calculate
2.
(i)
the angular velocity,
(ii)
the force towards the centre.
An object of mass 10 kg is whirled round a horizontal circle of radius 4 m by a revolving string inclined to the vertical. If the uniform speed of the object is 5 m s-1, calculate
3.
(i)
the tension in the string in N,
(ii)
the angle of inclination of the string to the vertical. X
A racing-car of 1000 kg moves round a banked track at a constant speed of 108 km h~ l . Assuming the total reaction at the wheels is normal to the track, and the horizontal radius of the track is 100 m, calculate the angle of inclination of the track to the horizontal and the reaction at the wheels.
4.
An object of mass 80 kg is whirled round in a vertical circle of radius 2m with a constant velocity of 6 ms_1 Calculate the maximum and minimum tensions in the string.
5.
Define the terms
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(a) acceleration, and (b) force. Show that the acceleration of a body moving in a circular path of radius r with uniform speed v is v 2/r, and draw a diagram to show the direction of the acceleration. 6.
A small body of mass m is attached to one end of a light inelastic string of length /. The other end of the string is fixed. The string is initially held taut and horizontal, and the body is then released. Find the values of the following quantities when the string reaches the vertical position : (a)
the kinetic energy of the body,
(b)
the velocity of the body,
(c)
the acceleration of the body, and
(d)
the tension in the string.
7.
Explain what is meant by angular velocity.
8.
Derive an expression for the force required to make a particle of mass m move in a circle of radius r with uniform angular velocity w.
9.
A stone of mass 500 g is attached to a string of length 50 cm which will break if the tension in it exceeds 20 N. The stone is whirled in a vertical circle, the axis of rotation being at a height of 100 cm above the ground. The angular speed is very slowly increased until the string breaks. In what position is this break most likely to occur, and at what angular speed? Where will the stone hit the ground?
10. A 2 kg ball on a string is rotated about a circle of radius 10 m. The maximum tension allowed in the string is 50 N. What is the maximum speed of the ball? 11. During the course of a turn, an automobile doubles its speed. How much additional frictional force must the tires provide if the car safely makes around the curve? 12. A satellite is said to be in geosynchronous orbit if it rotates around the earth once every day. For the earth, all satellites in geosynchronous orbit must rotate at a distance of 4.23×107 meters from the earth's center. What is the magnitude of the acceleration felt by a geosynchronous satellite? 13. A popular daredevil trick is to complete a vertical loop on a motorcycle. This trick is dangerous, however, because if the motorcycle does not travel with enough speed, the rider falls off the track before reaching the top of the loop. What is the minimum speed necessary for a rider to successfully go around a vertical loop of 10 meters? 14. During their physics field trip to the amusement park, Tyler and Maria took a rider on the Whirligig. The Whirligig ride consists of long swings which spin in a circle at rel atively high speeds. As part of their lab, Tyler and Maria estimate that the riders travel through a circle with a radius of 6.5 m and make one turn every 5.8 seconds. Determine the speed of the riders on the Whirligig.
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15. The tallest Ferris wheel in the world is located in Singapore. Standing 42 stories high and holding as many as 780 passengers, the Ferris wheel has a diamet er of 150 meters and takes approximately 30 minutes to make a full circle. Determine the speed of riders (in m/s and mi/hr) on the Singapore Flyer. ( GIVEN: 1.00 m/s= 2.24 mi/hr) 16. During the spin cycle of a washing machine, the clothes stick to the outer wall of the barrel as it spins at a rate as high as 1800 revolutions per minute. The radius of the barrel is 26 cm. a. Determine
the speed of the clothes (in m/s) which are located on the wall of the
spin barrel. b. Determine
the acceleration of the clothes.
17. Elmira, New York boasts of having the fastest car ousel ride in the world. The merry-go-round at Eldridge Park takes riders on a spin at 18 mi/hr (8.0 m/s). The radius of the circle about which the outside riders move is approximately 7.4 m. (a) Determine the time for outside riders to make one complete circle (b) Determine the acceleration of the riders. 18. A manufacturer of CD-ROM drives claims that the player can spin the disc as frequently as 1200 revolutions per minute. a. If
spinning at this rate, what is the speed of the outer row of data on the disc; this
row is located 5.6 cm from the center of the disc? b. What
is the acceleration of the outer row of data?
19. In the display window of the toy store at the local mall, a battery-powered plane is suspended from a string and flying in a horizontal circle. The 631-gram plane makes a complete circle every 2.15 seconds. The radius of the circle is 0.950 m. Determine the velocity of, acceleration of, and net force actin g upon the plane. 20. Dominic is the star discus thrower on South's varsit y track and field team. In last year's regional competition, Dominic whirled the 1.6 kg discus in a circle with a radius of 1.1 m, ultimately reaching a speed of 52 m/s before launch. Determine the net force acting upon the discus in the moments before launch. 21. Landon and Jocelyn are partners in pair figure skating. Last weekend, they perfected the death spiral element for inclusion in their upcoming competition. During this maneuver, Landon holds Jocelyn by the hand and swings her in a circle while she mai ntains her blades on the ice, stretched out in a nearly horizontal orientation. Determine the net force which must be applied to Jocelyn (m=51 kg) if her center of mass rotates in a circle with a radius of 61 cm once every 1.9 seconds.
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22. In an effort to rev up his class, Mr. H does a demonstration with a bucket of water tied to a 1.3-meter long string. The bucket and water have a mass of 1.8 kg. Mr. H whirls t he bucket in a vertical circle such that it has a speed of 3.9 m/s at the top of the loop and 6.4 m/s at the bottom of the loop. a. Determine
the acceleration of the bucket at each location.
b. Determine
the net force experienced by the bucket at each location.
c. Draw
a free body diagram for the bucket for each location and determine
the tension force in the string for the two locations. 23. Alexis is in her Toyota Camry and trying to make a turn off an expresswa y at 19.0 m/s. The turning radius of the level curve is 35.0 m. Her car has a mass of 1240 kg. Determine the acceleration, net force and minimum value of the coefficient of friction which is required to keep the car on the road. 24. Sheila (m=62 kg) is riding the Demon roller coaster ride. The t urning radius of the top of the loop is 12 m. Sheila is upside down at the top of the loop and experiencing a normal force which is one-half of her weight. Draw a fr ee body diagram and determine Sheila's speed. 25. In 2002, professional skateboarder Bob Burnquist became the first to succes sfully navigate a 360° full pipe turn. Determine the minimum speed which would be required at the top of the circular loop to make it through the 1.8-m radius pipe. 26. Justin is driving his 1500-kg Camaro through a horizontal curve on a level roadway at a speed of 23 m/s. The turning radius of the curve is 65 m. Determine the minimum value of the coefficient of friction which would be required to keep Justin's car on the curve. 27. A loop de loop track is built for a 938-kg car. It is a completely circular loop - 14.2 m tall at its highest point. The driver successfully completes the loop with an entry speed (at the bottom) of 22.1 m/s. a. Using
energy conservation, determine the speed of the car at the top of the loop.
b. Determine
the acceleration of the car at the top of the loop.
c. Determine
the normal force acting upon the car at the top of the loop.
28. Tyrone and Mia have masses of 84 kg and 59 kg respectively. They sit 1.0 m apart in the front center of Mr. H's Physics class. For some time, they each have been sensing a sort of electricity in their growing relationship. And now, six units into their Physics course, they have learned that they are gravitationally attracted to each other. Determine the magnitude of this force of gravitational attraction. 29. Determine the force of gravitational attraction between the Earth and the moon. Their masses are 5.98 x 10 24k g and 7.26 x 10
22
kg, respectively. The average distance separating
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the Earth and the moon is 3.84 x 10 8 m. Determine the force of gravitational attraction between the Earth and the moon. 30. Determine the force of gravitational attraction between the Earth and the sun. Their masses are 5.98 x 1024 kg and 1.99 x 10
30
kg, respectively. The average distance separating the
Earth and the sun is 1.50 x 10 11 m. Determine the force of gravitational attraction between the Earth and the sun. 31. Determine the acceleration of the moon about the Earth. (GIVEN: M Earth = 5.98 x 10 24 kg and Earth-moon distance = 3.84 x 10 8 m) 32. Determine the acceleration of the Earth about the sun. (GIVEN: Msun = 1.99 x 10
30
kg and
Earth-sun distance = 1.50 x 10 11 m) 33. Use Newton's law of gravitation to determine the acc eleration of an 85-kg astronaut on the International Space Station (ISS) when the ISS is at a height of 350 km above Earth's surface. The radius of the Earth is 6.37 x 10 6m . (GIVEN: M
Earth =
5.98 x 10
24
kg)
34. Determine the orbital speed of the International Space Sta tion - orbiting at 350 km above the surface of the Earth. The radius of the Earth is 6.37 x 10 6 m. (GIVEN: M Earth = 5.98 x 1024 kg) 35. Determine the orbital speed of the Earth as it orbits about the Sun. (GIVEN: M sun = 1.99 x 1030 kg and Earth-sun distance = 1.50 x 10
11
m)
36. Hercules is hoping to put a baseball in orbit by throwing it horizontally (tangent to the Earth) from the top of Mount Newton - 97 km above Earth's surface. With what speed must he throw the ball in order to put it into orbit? (GIVEN: M Earth = 5.98 x 10 24 kg; R Earth = 6.37 x 10 6 m) 37. Scientists determine the masses of planets by observing the effect of the gravitational f ield of those planets on nearby objects - mainly upon their moons. By measuring the orbital period and orbital radius of a moon about a planet, Newton's laws of motion can be used to determine the mass of the planet. Phobos, a moon of the planet Mars, was discovered in 1877. It's orbital radius is 9380 km and its orbital period is 0.319 days (2.77 x 10 4 seconds). Determine the mass of Mars based on this data. 38. Geostationary satellites are satellites which are orbiting the Earth above the equator and make one complete orbit every 24 hours. Because their orbital period is synchronized with the Earth's rotational period, a geostationary satellite can always be found in the same position in the sky relative to an observer on Earth. (GIVEN: MEarth = 5.98 x 10 24 kg) a. Determine
the orbital radius of a geostationary satellite.
b. Determine
the orbital speed of a geostationary satellite.
c. Determine
the acceleration of a geostationary satellite.
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39. In 2009, NASA's Messenger spacecraft became the second spacecraft to orbit the planet Mercury. The spacecraft orbited at a height of 125 miles above Mercury's surface. Determine the orbital speed and orbital period of Messenger. (GIVEN: R Mercury = 2.44 x 106 m; M Mercury = 3.30 x 10
23
kg; 1 mi = 1609 m)
40.
CIRCULAR MOTION | Gayan SK Perera
