Uniform Circular Motion

Uniform Circular Motion. From Fundamentals of Physics.

Take g=9.8ms-2, unless otherwise specified. Radius of Earth is approximately 6,370km

1.A satellite is in circular Earth orbit, at an altitude above the Earth’s surface. At that altitude, . What is the orbital speed, v, of the satellite?

2.What is the acceleration of a sprinter running at 10m/s when rounding a bend with a turn radius of 25m? To what does the acceleration vector point?

3.A sprinter runs at 9.2m/s around a circular track with a centripetal acceleration of 3.8m/s2. What is the track radius? How long does it take to go completely around the track at this speed?

4.If a remote controlled space probe can withstand the stresses of a 20g acceleration, what is the minimum turning radius of such a craft moving at a speed of one tenth the speed of light? How long would it take to make a 90 degree turn at this speed?

5.The fast train known as the TGV that runs from Paris in France has a scheduled average speed of 216km/h. If the train goes around a curve at that speed and the acceleration experienced by the passengers is limited to 0.050g, what is the smallest radius of curvature for the track that can be tolerated? If there is a curve of 1km radius, to what speed must the train be slowed to keep the acceleration below the limit?

6.An astronaut is rotated in a centrifuge at a radius of 5m. What is the astronaut’s speed if the acceleration is 7g? How many revolutions per minute are required to produce this acceleration?

7.What is the centripetal acceleration of an object on the Earth’s equator owing to the rotation of the Earth? What would the period of rotation of the Earth have to be in order that objects on the equator have a centripetal acceleration equal to 9.8m/s2?

8.A particle P travels with constant speed on a circle of radius 3m and completes one revolution in 20s. The particle passes through O (see below). The particle passes through O at t=0. Find the magnitude and direction of the following vectors. (a) With respect O, find the particle’s position vector at t=5, 7.5, and 10s. For the 5 second interval from the end of the fifth second to the end of the tenth second, find the particle’s (b) displacement and (c) average velocity. Find its (d) velocity and (e) acceleration at the beginning and end of that 5 second interval.

Uniform Circular Motion. From Fundamentals of Physics.

Take g=9.8ms-2, unless otherwise specified. Radius of Earth is approximately 6,370km

1.A satellite is in circular Earth orbit, at an altitude above the Earth’s surface. At that altitude, . What is the orbital speed, v, of the satellite?

Answer

Just use the formula, remember the radius is from the centre of Earth:

2.What is the acceleration of a sprinter running at 10m/s when rounding a bend with a turn radius of 25m? To what does the acceleration vector point?

Answer:

Again, a straight forward application of the formula:

And the acceleration vector points toward the centre of the circle.

3.A sprinter runs at 9.2m/s around a circular track with a centripetal acceleration of 3.8m/s2. What is the track radius? How long does it take to go completely around the track at this speed?

Answer:

That formula again:

And the time it takes is just:

Note how the algebra is done first to avoid rounding errors in the calculations.

Answer:

One tenth the speed of light is about ms-1, so:

90 degrees is just ¼ of a circle, so:

Answer:

Here we go again (care with units)…

And another rearrangement will give the maximum speed:

6.An astronaut is rotated in a centrifuge at a radius of 5m. What is the astronaut’s speed if the acceleration is 7g? How many revolutions per minute are required to produce this acceleration?

Answer:

Remember that frequency is the reciprocal of period so:

Or about 35 revolutions per minute.

Answer:

Interesting one this (again, care with units):

Bit messy, and the period, T is given by, so:

Or approximately 84 minutes.

a)This is just a bit of geometry really. At the particle has moved a quarter turn so the magnitude of the position vector (in blue) with respect to O is just and the direction is 45 degrees anticlockwise from x-axis. At the particle has moved on so that it is 3/8th around the circle. The magnitude, a, of the position vector is given by the cosine rule:

And the angle is 67.5 degrees. The last one is the easiest. It is just of magnitude 6m and direction 90 degrees.

b)It gets easier now, the displacement is just and the direction is 135 degrees.

c)So we have the distance covered from part (b) and the time taken is 5 seconds. So the velocity has magnitude 0.849ms-1 and the direction is 135 degrees again.

Continued….
d)The magnitude of the velocity is just:

And this does not change with time since the particle is travelling at a constant speed. The direction does change though. At the velocity vector is at 90 degrees and at it is at 180 degrees.

e)The magnitude of the acceleration is given by:

The directions are 180 degrees at t=5 and 270 degrees at t=10.