URGENT! help me wiv these questions...QUICK!! please :P

Well where are you stuck on?

Reply 2

hhhhhhhhhhppppp

OP

2

darkenergy

Well where are you stuck on?

all of it lol. i thought i cud do all of this but now ive jus got really confused wiv wot formulas to use. :confused:

im jus really really stuck wiv this. any help wud b really appreciated. anythin!! im desperate lol

Reply 3

well to start:
angular velocity is defined as radians per second. So you should start with the conversion.

Reply 4

lozzyjay

a grinding wheel of diameter 0.12m spins horizontally about a vertical axis. P is a typical grinding particle bonded to the edge of the wheel.
(A) If the rate of rotation is 1200 revolutions per minute, calculate
i) The angular velocity

Lauren xxx

ill see what i can do.....

a)....diameter= 0.12 m,
therefore, radius=0.12/2=0.06m
there are 1200 revolutions in 60 seconds, therefore revolutions per sec (or frequency)= 1200/60=20 revs/sec or Hz

now use:
ω=2 PI f (i oculdnt find the symbol for pi anywhere)
=2.pi.20
=125.67 rads/sec thats ur angular velocity

ill post the rest in a sec! :smile:

Reply 5

steve2005

17

lozzyjay

a grinding wheel of diameter 0.12m spins horizontally about a vertical axis. P is a typical grinding particle bonded to the edge of the wheel.
(A) If the rate of rotation is 1200 revolutions per minute, calculate
i) The angular velocity

Can you post the formulas you have been given ? Then I might be able to help......
http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html

Picture 7.png72.4KB

Reply 6

lozzyjay

a grinding wheel of diameter 0.12m spins horizontally about a vertical axis. P is a typical grinding particle bonded to the edge of the wheel.
(A) If the rate of rotation is 1200 revolutions per minute, calculate

ii)The acceleration of P

and now we know that ω=125.67 rads/sec

to calcuate the acceleration, use,
a=ω² r
(we converted the diameter to radius earlier on)
a=125.67² x 0.06
= 947.48 ms^-2

Reply 7

lozzyjay

a grinding wheel of diameter 0.12m spins horizontally about a vertical axis. P is a typical grinding particle bonded to the edge of the wheel.
(A) If the rate of rotation is 1200 revolutions per minute, calculate

iii)the magnitude of the force acting on P if its mass is 1.0x10^-4

and now we know the acceleration is 947.48m/s²

force=?

use

f=ma
=1.0x10^-4 x 947.48
=9.5x10^-2 N....this is the force acting on P

Reply 8

Yes.

Reply 9

lozzyjay

a grinding wheel of diameter 0.12m spins horizontally about a vertical axis. P is a typical grinding particle bonded to the edge of the wheel.
(A) If the rate of rotation is 1200 revolutions per minute, calculate
i) The angular velocity
ii)The acceleration of P
iii)the magnitude of the force acting on P if its mass is 1.0x10^-4
(B) The maximum radial force at which P remains bonded to the wheel is 2.5N.
i)calculate the angular velocity at which P will leave the wheel if its rate of rotation is increased.

im not sure about the bits here onwards, but ill try.....

so maximum F=2.5N, and we know m for P, lets use

F=ma
rearranging,
a=F/m
a=2.5/0.0001
max a= 25000 N

and now we know a, we also know the radius which is 0.06m, use
a=ω² r
ω²=a/r
=25000/0.06
ω=√ 25000/0.06
= 645.5 rads/sec.......this is the new angular velocity

Reply 10

lozzyjay

a grinding wheel of diameter 0.12m spins horizontally about a vertical axis. P is a typical grinding particle bonded to the edge of the wheel.
(A) If the rate of rotation is 1200 revolutions per minute, calculate
i) The angular velocity
ii)The acceleration of P
iii)the magnitude of the force acting on P if its mass is 1.0x10^-4
(B) The maximum radial force at which P remains bonded to the wheel is 2.5N.
i)calculate the angular velocity at which P will leave the wheel if its rate of rotation is increased.
(C) If the wheel exceeds this maximum rate of rotation, what will be the speed and direction of motion of particle P immediately after it leaves the wheel?

right, in this last bit, i think theyre asking u to find the linear velocity, its just differently worded

so, we know that ω= 645.5 rads/sec, and that a=25000N,and also that the radius is 0.06 m,
use

a= v² /r
25000x0.06=v²
therefore the linear vel=38.73 m/s

and thats the end of the question!

hope its been of help!.... :smile:

Reply 11

Direction is perpendicular to the centripetal force.

Reply 12

steve2005

17

PSdilemma

right, in this last bit, i think theyre asking u to find the linear velocity, its just differently worded

so, we know that ?= 645.5 rads/sec, and that a=25000N,and also that the radius is 0.06 m,
use

a= v² /r
25000x0.06=v²
therefore the linear vel=38.73 m/s

and thats the end of the question!

hope its been of help!.... :smile:

Impressive BUT what is the direction of the motion

Reply 13

steve2005

Impressive BUT what is the direction of the motion

oops! forgot to answer that! :biggrin:

.....hmmmm, all i can say is, from what information is given, the direction would be tangential to the direction of travel just before it leaves the wheel....(because the particle now leaves the wheel, there is no centripetal force acting on it anymore, it will therefore travel in a straight line, ie, at a tangent to the direction it had just before leaving)

Reply 14

steve2005

17

PSdilemma

and now we know that ?=125.67 rads/sec

to calcuate the acceleration, use,
a=?² r
(we converted the diameter to radius earlier on)
a=125.67² . 0.006
= 947.48 ms^-2

The 947.48 is found ...a=125.67² . 0.06
= 947.48 ms^-2

i.e the radius is 0.06 and not 0.006. In the calculations you used the correct figure... but you wrote down the wrong figure.

Very impressive work and done very fast.

Reply 15

darkenergy

Direction is perpendicular to the centripetal force.

there! it was answered already!! :smile:

....i didnt see it! :rolleyes:

.....and but ofcourse, this is a more accurate answer! :smile:

Reply 16