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Circular motion question with weight

http://s28.postimg.org/naff7dcl9/phy_problem_1.png

I can't understand these kind of problems, when we use arrows etc. and find the overall force..

I pictured the car, with two forces vertically downwards: mg and cf force (mv^2/r)

and one force upwards, which is the reaction force, however I am kind of confused about the reaction force right now, I thought it was the force of same magnitude but the opposite direction, so I pictured another mg force coming vertically upwards.

The answer to this is D and I don't have a good idea why.
Original post by The Best1
http://s28.postimg.org/naff7dcl9/phy_problem_1.png

I can't understand these kind of problems, when we use arrows etc. and find the overall force..

I pictured the car, with two forces vertically downwards: mg and cf force (mv^2/r)

and one force upwards, which is the reaction force, however I am kind of confused about the reaction force right now, I thought it was the force of same magnitude but the opposite direction, so I pictured another mg force coming vertically upwards.

The answer to this is D and I don't have a good idea why.


There are two forces on the car
* its weight, mg (downwards)
* reaction from the road, R (upwards)

Together, the resultant of these two must equal the centripetal force on the car (downwards) mv2/r

So if we say downwards is positive

mg - R = mv2/r
Reply 2
Avatar for lerjj
lerjj
13
Just to clarify- for circular motion the resultant force equals mv^2/r. So you know the weight is mg, and you know the weight plus the reaction force=mv^2/r. Therefore the reaction force is the difference between them, or mg-mv^2/r.
Reply 3
Avatar for Phichi
Phichi
11
Original post by lerjj
the resultant force equals mv^2/r


Towards the center of the of the circle to clarify for OP as he mentioned 'cf' as mv2r\dfrac{mv^2}{r}


Original post by The Best1


Just to clarify as above, the force given by mv2r\dfrac{mv^2}{r} is the centripetal force, which is directed towards the center of the circle. You mentioned 'cf', which makes me think you are referring to centrifugal force, which is wrong, it is a common misconception. Check these out:

http://en.wikipedia.org/wiki/Centrifugal_force

http://en.wikipedia.org/wiki/Centripetal_force
Just to clarify even further.
The OP stated that the force (which he referred to as "cf") was "vertically downwards" and equal to mv2r \frac{mv^2}{r}
The infamous and non existent centrifugal force would have been directed upwards.

lerjj
You haven't clarified this for the OP at all
You stated that
".. the weight plus the reaction force=mv^2/r"

This gives
mg + R = mv2/r

and does not produce answer D
It actually gives answer C.
Reply 5
Avatar for lerjj
lerjj
13
Original post by Stonebridge
Just to clarify even further.
The OP stated that the force (which he referred to as "cf") was "vertically downwards" and equal to mv2r \frac{mv^2}{r}
The infamous and non existent centrifugal force would have been directed upwards.

lerjj
You haven't clarified this for the OP at all
You stated that
".. the weight plus the reaction force=mv^2/r"

This gives
mg + R = mv2/r

and does not produce answer D
It actually gives answer C.

Sorry, can't do vector arrows... I did intend that statement to take signs into account. So (-mg)+(+N)=(-mv^2/r), taking +ve to mean up. Adding +mg to both sides gives D, which was my point when I said resultant force. I shouldn't have been lazy with the signs though, that probably wasn't helpful

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