Q.circular motion (1 Viewer)

[freaking_out]

attached is the question....btw, is this in ze course?

 

[MyLuv]

I 've seen this1 in Year10Physics in my country
Angular velocity of the object: w= 2*2(pi)/1=4(pi)
Friction play the role of centripetal force in this case; and the greatest value for friction is Mmg (ortherwise it will slip out of the disc):
Fc=mw2r=Mmg
--> m*16*(pi)2*r=Mmg
-->r=Mg/16*(pi)2~0.06M metres
--> max distance r~ 6M cm

 

[freaking_out]

hey, thanks for the reply, i get the mathematics, but i don't get the actual question/concept...

can u explain to me why, the "friction plays the role of centripetal force"?? and can u also explain to me this question in general?

 

[MyLuv]

In general, when U have a circular motion ,U need a centripetal force and in this case there 's no others than vertically gravity,and reaction force so Friction must be centripetal
(Actually,when the object occupy circular motion it'll tend to fall out of the disc,'n the friction resist this motion thus acting as centripetal force )

 

[freaking_out]

hmm...the answer at the book says the following which makes me more confused:

"the frictional force keeping object on the table is Mmg. The object will remain on the table if:

centripetal force<=friction

they, then take the above inequality and work out:
r<= 6Mcm


then it concludes:
"so the object can be placed nearly 6Mcm from the centre of
the disc in order to stay on it."

can someone explain?

 

[ND]

Originally posted by freaking_out
hmm...the answer at the book says the following which makes me more confused:

"the frictional force keeping object on the table is Mmg. The object will remain on the table if:

centripetal force<=friction

they, then take the above inequality and work out:
r<= 6Mcm


then it concludes:
"so the object can be placed nearly 6Mcm from the centre of
the disc in order to stay on it."

can someone explain?

Which part is it that you don't understand? Is it about the centripetal force <= friction? If so, think about it, the only thing keeping the object in the table is the fritional force. If the frictional force is not great enough to hold on, the circular motion will push it off.

edit: haha you went to all the trouble of scanning it and uploading it just so you don't have ot write like four lines?

edit 2: i should probabliy rephrase that.

 

[freaking_out]

Originally posted by ND
Which part is it that you don't understand? Is it about the centripetal force <= friction? If so, think about it, the only thing keeping the object in the table is the fritional force. The centripetal force wants to push the object on the table, and will do so if teh frictional force is not great enough to 'hold on'.

edit: haha you went to all the trouble of scanning it and uploading it just so you don't have ot write like four lines?

but technically speaking, doesn't cetripetal force try to pull the mass towards the centre?? if so then how is it gonna fall off the "circular disk" in the first place....or is this getting into centrigual force, in where particles get thrown out?

 

[ND]

Originally posted by freaking_out
but technically speaking, doesn't cetripetal force try to pull the mass towards the centre?? if so then how is it gonna fall off the "circular disk" in the first place....or is this getting into centrigual force, in where particles get thrown out?

Read my edit, rephrased. Yeh i meant to say centrifugal force instead of centripetal, i think the book did too. Btw centrifugal force = mv^2/r too.