Questions on Circular Motion


by asteg123
Tags: circular, motion
asteg123
asteg123 is offline
#1
Sep9-06, 10:40 PM
P: 13
1. When the radius and mass are held constant, what do you expect to happen to the centripetal force if the frequency of rotation is increased?

2. What will happen if a rotating platform was set up slightly off the horizontal? Would anything change? What in particular?

3. A ball rotates in a horizontal circle at a constant speed of 10m/s. what are the tensions in the upper and lower strings? The mass of the ball is 3kg.


4. What angle of bank is necessary for a car to make it around a 130m curve at a speed of 60kph without relying on friction.

5. Determine the speed of the satellite orbiting at a height of 700km above the Earth's surface.
mass of earth = 5.98 x 10^24 kg
radius of earth = 6.38 x 10^6 m


~~~~~~~~~~~~~~~~~~~~~~~~

I would greatly appreciate the help...

Oh, and btw, I also need Free Body Diagrams for problems 3 - 5 (so that i won't get confused )

Thanks again....
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mbrmbrg
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#2
Sep9-06, 10:52 PM
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P: 497
asteg, what are your thoughts so far?
Especially about the free body diagram--have you come up with anything at all?
asteg123
asteg123 is offline
#3
Sep9-06, 11:26 PM
P: 13
So far, I had only done number 4, but i am not very sure of what i did...

I had made a FBD of this and it showed a vector at a certain angle (WRT the y-axis) and towards the right... this vector represents the Normal force... there is also a vector downwards, and it is W=mg

and from there, i'm stuck...


With regards to the other problems, I always get stuck on the FBD, so I can't really figure the problem out...

Hope you could help me...

asteg123
asteg123 is offline
#4
Sep10-06, 01:43 AM
P: 13

Questions on Circular Motion


Ok, so far, I had answers to number 3 & 4, However, I'm not sure if they're right...

Here's what i did, on number 3, I made an FBD and here it is...

Can anyone verify this.... This is my equation

Letting the angle be A

the Summation Fy=0=T'sinA - T''sinA-w
the summation Fx=m*a=(mv^2)/R=T''cosA+T'cosA

then i simply did algebra and got the answer

T'' = 325.62N and T' = 360.9N

~~~ In need of verification ~~~

and for number 4

where the angle is A

i got the summation Fx=m*a=n*sinA where a is a-rad
and summation Fy=0=n*cosA-m*g
I have yet to solve this, and i had noticed the lot of conversion of units in this problem

However, i still need to verify if what I did was correct...

~~ in need of verification ~~

and with regards to 1, 2 and 5 , I still have no answers... especially in the numbers 1 & 2 (I'm bad at theory)

Anyways... hopefully, someone will provide help....

Thanks in advance...
radou
radou is offline
#5
Sep10-06, 04:17 AM
HW Helper
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P: 3,225
Quote Quote by asteg123
1. When the radius and mass are held constant, what do you expect to happen to the centripetal force if the frequency of rotation is increased?
The centripetal force increases, since [tex]F_{cp}=m\omega^2r[/tex], where [tex]\omega[/tex] is the angular speed of the ball (frequency of rotation).

Quote Quote by asteg123
5. Determine the speed of the satellite orbiting at a height of 700km above the Earth's surface.
mass of earth = 5.98 x 10^24 kg
radius of earth = 6.38 x 10^6 m
You have to set the centripetal force and the gravitational force to equal, which is a condition for a satellite to orbit around the Earth.
asteg123
asteg123 is offline
#6
Sep10-06, 05:36 AM
P: 13
Quote Quote by radou
You have to set the centripetal force and the gravitational force to equal, which is a condition for a satellite to orbit around the Earth.
Uhmm... is the gravitational force the constant of GM earth??

Is there any other way of solving this other than this method... coz we still haven't discussed GM yet...

I think the method is somewhat involving frequency..
I'm not really sure...
Delzac
Delzac is offline
#7
Sep10-06, 05:51 AM
P: 389
well i'm not very sure it Frequency can be used here, but what is the point of giving u the mass of the earth if Gravitational Force will not be used.
asteg123
asteg123 is offline
#8
Sep10-06, 06:10 AM
P: 13
hmm... true...

I guess I will use the Gravitational force...

THANKS!!!


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