QUESTIONS
Questions
1
A particle P is moving in a circle with uniform
speed. Draw a diagram to show the direction of
the acceleration a and velocity v of the particle
at one instant of time.
5
The Singapore Flyer is a large Ferris wheel
of radius 85 m that rotates once every
30 minutes.
P
2
State what provides the centripetal force that
causes a car to go round a bend.
3
State the centripetal force that acts on a particle
of mass m when it is travelling with linear
speed v along the arc of a circle of radius r.
4
(IB) At time t = 0 a car moves off from rest in a
straight line. Oil drips from the engine of the car
with one drop every 0.80 s. The position of the
oil drops on the road are drawn to scale on the
grid below such that 1.0 cm represents 4.0 m.
The grid starts at time t = 0.
a) Calculate the linear speed of a point on the
rim of the wheel of the Flyer.
b) (i) Determine the fractional change in the
weight of a passenger on the Flyer at
the top of the ride.
(ii) Explain whether the passenger has a
larger or smaller apparent weight at the
top of the ride.
c) The capsules need to rotate to keep the
floor of the cabin in the correct place.
Calculate the angular speed of the capsule
about its central axis.
Direction of motion
1.0 cm
6
a) (i) State the feature of the diagram that
indicates that the car accelerates at the
start of the motion.
a) Quito in Ecuador (14 minutes of arc south
of the Equator)
(ii) Determine the distance moved by the
car during the first 5.6 s of its motion.
b) The car then turns a corner at constant
speed. Passengers in the car who were
sitting upright feel as if their upper bodies
are being “thrown outwards”.
(i) Identify the force acting on the car, and
its line of action, that enables the car to
turn the corner.
(ii) Explain why the passengers feel as if
they are being thrown outwards.
The radius of the Earth is 6400 km. Determine
the linear speed of a point on the ground at the
following places on Earth:
b) Geneva in Switzerland (46° north of the
Equator)
c) the South Pole.
7
A school bus of total mass 6500 kg is carrying
some children to school.
a) During the journey the bus needs to travel
round in a horizontal curve of radius 150 m.
The dynamic coefficient of friction between
the tyres and the road surface is 0.7.
Estimate the maximum speed at which the
driver should attempt the turn.
265
6
CIR C UL A R MOT ION A N D GRAV ITATION
b) Later in the journey the driver needs to
drive across a curved bridge with a radius of
curvature of 75 m. Estimate the maximum
speed if the bus is to remain in contact with
the road.
8
A velodrome used for bicycle races has a
banking angle that varies continuously from 0°
to 60°. Explain how the racing cyclists use this
variation in angle to their advantage in a race.
11 Determine the distance from the centre of the
Earth to the point at which the gravitational field
strength of the Earth equals that of the Moon.
12 The ocean tides on the Earth are caused by the
tidal attraction of the Moon and the Sun on the
water in the oceans.
a) Calculate the force that acts on 1 kg of
water at the surface of the sea due its
attraction by the
Data needed for these questions:
(i) Moon
Radius of Earth = 6.4 Mm;
Mass of Earth = 6.0 × 1024 kg;
Mass of Moon = 7.3 × 1022 kg;
Mass of Sun = 2.0 × 1030 kg;
Earth–Moon distance = 3.8 × 108 m;
Sun–Earth distance = 1.5 × 1011 m;
G = 6.67 × 10–11 N m2 kg–2
(ii) Sun.
9
Deduce how the radius R of the circular orbit of
a planet around a star of mass ms relates to the
period T of the orbit.
10 A satellite orbits the Earth at constant speed as
shown below.
satellite
b) Optional – difficult. Explain why there are
two tides every day at many coastal points
on the Earth.
[Hint: there are two parts to the answer,
why a tide at all, and why two every day.]
13 There are two types of communication
satellite. One type of communication satellite
orbits over the poles at a distance from
the centre of the Earth of 7400 km; the
other type is geostationary with an orbital
radius of 36 000 km. Geostationary satellites
stay above one point on the equator whereas
polar-orbit satellites have an orbital time of
100 minutes.
Calculate:
a) the gravitational field strength at the
position of the polar-orbit satellite
Earth
b) the angular speed of a satellite in
geostationary orbit
c) the centripetal force acting on a
geostationary satellite of mass 1.8 × 103 kg.
a) Explain why, although the speed of
the satellite is constant, the satellite is
accelerating.
b) Discuss whether or not the gravitational
force does work on the satellite.
266
E N D - O F -TO P I C Q U E S T I O N S
Solutions for Topic 6 – Circular motion and gravitation
ν
1.
α
P
2. friction between the tyres and the road.
mv2 .
3. F = _
r
4. a) (i) The drops are increasingly far apart and so the speed is increasing.
(ii) 5.6 s is 6 time intervals so the distance travelled is 14.4 cm on the scale, or 57.6 m on
the ground.
b) (i) centripetal force; acting towards the centre of the circle
(ii) Passengers are in a rotating frame of reference. Seen from above the passengers would
move in a straight line from Newton’s First Law of motion but friction acts at seat to provide
centripetal force to centre of circle. Passenger interprets the reaction to this force
as being flung outwards.
534 = 0.297 m s–1.
5. a) Circumference is 2π × 85 = 534 m. So linear speed is _
30 × 60
v2
_
m
2
v so fractional change = _
r
v2 _
0.592
_
–4
b) (i) change in weight = m _
r
mg = gr = 9.8 × 170 = 2.1 × 10
(ii) a smaller apparent weight as in passenger frame of reference there is an apparent additional
upward force
c) Capsule turns 2π rad in 30 minutes, so 3.5 mrad s–1.
2π
6. Angular speed of Earth = __
= 7.3 × 10–5 rad s–1
24 × 60 × 60
Linear speed v = ω r cos θ where θ is the latitude.
a) 14’ of arc = 4.1 mrad, linear speed = 7.3 × 10–5 × 6.4 × 106 × 1 = 470 m s–1
b) 46° gives 320 m s–1
c) At the geographical south pole the linear speed is zero.
7. a) Maximum friction force = 6500 × 9.8 × 0.7 = 44.6 kN ___________
v2 = _
6500 v2. So v = __
44600 × 150 = 32 m s–1
Centripetal force required = m _
max
r
150
6500
2
_______
__
v
_
–1
b) m r < mg; v > √ rg . Maximum speed = √75 × 9.8 = 27 m s
√
8. The component of the normal reaction force acting horizontally contributes to the centripetal force
so the faster the cyclist is travelling, the greater the component required and this is achieved by
moving up the slope to a point where the slope angle is greater.
9. Gravitational force on planet provides the centripetal force on the planet (Keplar’s third law)
mS mp
so mp ω2R = G _
R2
m
re-arranging ω2 = G _3S
R
2π
ω = 2πf = _
T
2 3
4π
R
and T2 = _
Gms
© Oxford University Press 2014: this may be reproduced for class use solely for the purchaser’s institute
1
E N D - O F -TO P I C Q U E S T I O N S
10. a) Speed is a scalar but velocity is a vector. The direction of the velocity is constantly changing, so
the vector velocity is changing too. Acceleration occurs when velocity changes and so there is
acceleration in this case.
b) Work done = distance travelled × force in direction of distance travelled. The force
(acceleration) acting and the distance travelled are at 90° to each other so in this case no work
is done. OR
Work is done when kinetic energy or potential energy change. The speed is constant so kinetic
energy is constant. Distance from Earth is constant so gravitational potential energy does not
change. So no work is done.
Gm
11. g = _
r2
GmE
GmM
=
2
rE
rM2
Where rE and rM are the distances from the centres of Earth and Moon respectively to the point
where the field strengths are equal (known as the Lagrangian point).
rE2
mE _
6 × 1024 = 82
So r 2 = _
=
mM
7.3 × 1022
M
rE
__
√
rM = 82 = 9.06
9
Therefore the point is th of the way from the Earth to the Moon (3.42 × 108 m).
10
G mM
12. a) (i) force on 1 kg of water = _
= 3.4 × 10–5 N due to Moon
r2
G mS
(ii) force on 1 kg of water = _
= 6.0 × 10–3 N due to Sun
r2
b) When the Moon is overhead there is a gravitational force of attraction (a tide) on objects. So
fluids are able to respond to this by an increase in the water level (tides are also observed in
the rocks). There are two tides because there is a corresponding bulge in the surface on the
opposite side of the Earth.
GM = __
6.7 × 10–11 × 6 × 1024 = 7.3 N kg–1
13. a) g = – _
r2
(7.4 × 106)2
2π =
b) The satellite orbits in 24 hours; orbital time = 86400 s. Angular speed = _
86400
7.3 × 10–5 rad s–1
–11
24
3
GME m ____
mv2 = _
c) _
= 6.7 × 10 × 6 × 10 7 2× 1.8 × 10 = 560 N
2
r
r
(3.6 × 10 )
© Oxford University Press 2014: this may be reproduced for class use solely for the purchaser’s institute
2