• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

circular motion (1 Viewer)

elizabethy

I'm the best !!
Joined
Oct 23, 2002
Messages
1,160
Location
Sydney
Gender
Female
HSC
2003
hi

i seriously need helppppppp on my prac related to circular motion

can someone help me answer thw question on page 55 JACARANDA

plzzz have a look otherwise i will fail!!!!!!!!!
 

Jago

el oh el donkaments
Joined
Feb 21, 2004
Messages
3,691
Gender
Male
HSC
2005
if you actually type up/scan the question you might get a better response.
 

clever angel

Member
Joined
Aug 24, 2003
Messages
240
Location
sydney
Gender
Female
HSC
2005
these r the questions , see if ne one can help her?

what does the slope of your v squared versus graph represent?

what role does gravity play in this experiment?


actually even i have problems with the same prac, but no one replied in my thread
 

KFunk

Psychic refugee
Joined
Sep 19, 2004
Messages
3,323
Location
Sydney
Gender
Male
HSC
2005
I'll explain them in reverse because it works well that way.

Gravity is providing you with the force that you are trying to balance with centripetal force. You are trying to keep the hanging mass still. When it is still then F<sub>c</sub> = F<sub>g</sub>. Since F<sub>g</sub> is a constant (for a given hanging mass) then every time you balance the same hanging mass you are creating the same centripetal force... this is where the graph comes into it.

If you look at your graph for, say 100g, you will notice that as v<sup>2</sup> increases, so does r. F<sub>c</sub>=mv<sup>2</sup>/r. The mass of the orbiting bob is a constant and so it does not come into play. You will notice from the equation that if you increase the radius then in order to create the same centripetal force v<sup>2</sup> must increase. The slope then represents the centripetal acceleration required to generate a centripetal force to balance the downward force of gravity (remembering F=ma and that a<sub>c</sub>= v<sup>2</sup>/r).

I hope I explained that clearly enough.

EDIT: Bugger, I can't make the sub/superscripts work... are they turned off in the physics forum?
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top