# Planetary motion problem

## Homework Statement

a planet with radius of 12km spins at 520revs/s
find:
a) avg speed of a point on the planets equator over 2.5 of a revolution
b)find avg acceleration on stars circumfrance over 3/4 of a rev
c)find distance covered by point on the equator in 1 second
d) find displaement

f=520rev/s r=12000m

fc=mv^2/r
T=1/f
v=2piR/T

## The Attempt at a Solution

a) if i construct a vector triangle using v1 and v2 at the beginning of 3/2 of a rev and at the end, i can believe i can solve for vavg this way making a triangle where Vavg opposes the angle of 144 degrees...
v=2piR/T
where T=1/f

so i get v=2Pi*Rf

i considered V1and V2 = v

vavg^2 = 2(2Pi*R*f)^2-2(2Pi*R)^2cos144

is this okay?

b)m*ac=mv^2/R
where ac = centripetal accell
and v = 2pi*R*F

ac = ((2pi*R)*f)^2/R

c)m*ac= (d/t)^2/R

if t=1s

d=√(R*ac)

d)... i think i may have solved displacement above.. perhaps i am confusing the distance and displacement.

## Answers and Replies

a) avg speed of a point on the planets equator over 2.5 of a revolution

Here, avg speed is asked not avg velocity and in this case avg speed will be same as instantaneous speed.

b)find avg acceleration on stars circumfrance over 3/4 of a rev

avg acceleration is change in velocity over time.

ac = ((2pi*R)*f)^2/R

this is instantaneous acceleration, if all the units are in standard form. f should be in radians per second.

c)m*ac= (d/t)^2/R

if t=1s

d=√(R*ac)

seems great. d)... i think i may have solved displacement above.. perhaps i am confusing the distance and displacement.

distance is the path traversed, that is the curve along which you move (the object, here point on the equator). displacement is the length of the straight line connecting initial and final position. moreover, displacement is a vector but distance is not.

Here, avg speed is asked not avg velocity and in this case avg speed will be same as instantaneous speed.

avg acceleration is change in velocity over time.

so since its 1/3 of the distance

V=D/T => V (D/3)/T =(2PiR*f)/3

this is instantaneous acceleration said:
f[/I] should be in radians per second.

units are wanted in m/s^2

M*ac = V^2/R
ac = (((D/(3/4))^2/T)/R

distance is the path traversed said:
So what can i do to find the displacement ?

so since its 1/3 of the distance

V=D/T => V (D/3)/T =(2PiR*f)/3

what is 1/3 of the distance?!?! anyways, its still not right.

M*ac = V^2/R

its not the correct expression. the expression is M*ac =M* V^2/R

But the acceleration in this equation is instantaneous. You require Average Acceleration.

Average Acceleration = ((final velocity)-(initial velocity))/(time interval)

Use vector addition/subtraction for calculating it.

So what can i do to find the displacement ?

check the image to be clear. the chord represents displacement and the arc represents distance.

calculate Δθ from the given data. Then a bit of trigonometry should give you the answer.

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a) vavg = 2*pi**R*F

cause velocity is the same at every point in circular motion

b) ac = V^2(avg)/R

d)Δθ=ΔS/R

where S is my distance from C (S2-S1)/R = S2/R

and then use cosine law with the angle.. Also was a type for the formula mac=mv^2/R , it was indicated in the beginning.

a) vavg = 2*pi**R*F

cause velocity is the same at every point in circular motion

Yes exactly. b) ac = V^2(avg)/R

this is not the correct expression. you need to use vectors to solve this.
Where would the velocity vector be pointing before and after 3/4th of revolution? What will be their magnitude?
draw it on a piece of paper, that should help
remember Average Acceleration = ((final velocity)-(initial velocity))/(time interval)

Subtract them vectorially and find time from the given info.
d)Δθ=ΔS/R

where S is my distance from C (S2-S1)/R = S2/R

what is C ?.....center point of circle??
the formula is right. calculate the distance just like you previously did and that distance is ΔS.
and then use cosine law with the angle.. Also was a type for the formula mac=mv^2/R , it was indicated in the beginning.
yes, that should solve it for you.