Wider Applications Of Circular Motion

Question 1

A student looking at a = v2/r concludes that centripetal acceleration decreases as the distance r increases because r appears in the denominator of the equation.
Another student looking at a = rω2 reasons that the centripetal acceleration increases with increasing r because in this equation r appears in the numerator.
Who is right?

Answer 1

Both students are partly correct although neither is telling the whole story. What the first student should say is, ‘If a = v^2/r then a will decrease with increasing r provided v remains constant.’ The second student should say ‘If a = rw^2, a will increase with increasing r provided w remains constant.’ The statements are not contradictory as it is impossible for both v and w to remain constant if r is increasing. This follows from the relationship v = rw. If v is to remain constant as r increases, then w must decrease proportionately. Similarly if w is to remain constant, v has to increase.

Question 2

An electron is moving at 1% of the speed of light in a circle of radius 1.4 m in a particle accelerator. Calculate the force the magnetic field is exerting on the electron, and state the direction of the force.

Answer 2

5.86 x 10⁻¹⁸ N

Question 3

A single stationary particle decays into a proton and an antiproton (a particle which has the same mass as a proton), which move in curved tracks of radius 5.06 cm away from each other. If the force on each particle is 5 x 10-10 N, calculate the speed each particle is moving at.

Answer 3

1.23 x 10⁸ ms⁻¹

Question 4

An electron is accelerated to a velocity of 2.1 x 107 ms-1 by a particle accelerator, in a magnetic field of strength B = 0.55 mT. Calculate the radius of the electron’s path. Use the equation F = Bqv to calculate the force on the electron due to the field.

Answer 4

0.217 m

Question 5

A planet orbiting a star experiences a force of magnitude 3.0 × 10^22 N due to the gravitational attraction of the star. If the planet has a speed of 2.0 × 10^5 ms−1 and it takes 2.6 × 10^7 s to complete one orbit, calculate the mass of the planet and the radius of its orbit.

Answer 5

mass = 6.3 x 10^23 kg
radius = 8.3 x 10^11 m

Question 6

The radius of the Earth is 6.38 × 10^6 m and that the radius of the Earth’s orbit around the Sun is 1.50 × 10^11 m. Calculate both the speed and angular speed for:
a point on the Earth’s equator, due only to the rotation of the Earth on its axis,
the centre of the Earth, due to the motion of the Earth around the Sun

Answer 6

a. ω = 7.27 x 10^-5 rad s^-1
v = 4.64 x 10^2 ms^-1
b. ω = 1.99 x 10^-7 rad s^-1
v = 2.99 x 10^4 ms^-1

Question 7

Calculate the speed of a satellite in circular orbit 1000 km above the Earth’s surface. Take the mass of the Earth as 5.98 × 10^24 kg and the radius of the Earth as 6.38 × 10^6 m.

Answer 7

7.35 x 10^3 ms^-1

Question 8

It is common knowledge that taking a bend too fast can cause a car to skid and the tighter the bend the more likely it is that skidding will occur. For a bend on a motorway to be ‘safe’, the curve must be sufficiently gentle that a car will not skid when travelling at the maximum allowable speed of 70 mph..
In the case of a car on a horizontal road surface, the maximum friction acting is Fmax = µW where Fmax is the magnitude of the maximum frictional force, W is the magnitude of the car’s weight and µ is a constant called the coefficient of static friction. The value of µ depends on the nature of the surfaces: an average value may be µ = 0.70; for poor tyres on a wet road surface µ = 0.30.
What is the tightest (smallest radius) safe bend?

Answer 8

F = µW
W = mg
F = µmg
F = µmg = (mv^2)/r = m(v^2/r)
F = (rF/m)^0.5 = (rµmg/m)^0.5 = (rµg)^0.5
vr = mv^2/F = mv^2/µmg = v^2/µg

It is important to have a safety margin because if cars travelling over 71mph or with used tyres crash the road is unsafe.

µ = 0.3
v = 40ms^-1 or 90mph

r = 5.4 x 10^2

Question 9

Calculate the maximum constant speed with which a car can take a bend with radius of curvature 3.0 × 102 m without skidding, if the value of µ is 0.40.

Answer 9

34 ms”-1

Question 10

Mars has a mass of 6.42 × 1023 kg and rotates with a period of 1.03 Earth days. At what distance from the centre of Mars should a satellite be put into orbit so that it remains located above a fixed point on the Martian surface?

Explain how your answer to Q1 would change if the mass of the satellite was doubled

Answer 10

2.05 x 10^7 m

No change: the mass of the satellite has no effect on the calculation

Question 11

The strength of the gravitational force experienced by Io, one of Jupiter’s moons, is 5.19 × 1022 N. If it has a period of orbit of 1.77 Earth days and its distance from the centre of Jupiter is 4.22 × 108 m, calculate its mass.
Explain why a satellite cannot be put into a geostationary orbit directly over London.

Answer 11

7.29 x 10^22 kg
The centripetal force holding a satellite in circular orbit must act towards the centre of the circle that the satellite in moving in.
The centripetal force in this case is provided by the Earth’s gravitational field, which acts from the centre of mass, which is the centre of the Earth.
The satellite must therefore be in orbit in a plane containing the centre of the Earth; for a geostationary satellite, this must be directly above the equator.