Question

Can anyone explain how to do this problem:
A grinding wheel .29m in diameter spins up from rest to 2705 rpm in 2 seconds
a- Calculate its final angular velocity in rad/s
b- what is the angular acceleration of the grinding wheel
c- what is the linear speed of a point on the edge of the grinding wheel after 2 seconds
d- what is the total linear acceleration of a point on the edge of the grinding wheel after 2 seconds
e- if the grinding wheel has a mass of 3kg and can be considered as a uniform soild disk what is the magnitude of the torque acting on the grinding wheel's angular mome

Answer 1

\[\frac{ rev.s }{ \min} *\frac{ 1 \min }{ 60 \sec } * \frac{ 2\pi rad.s }{ rev} =\frac{ rad.s }{ \sec }\]

Answer 2

can you do 'a' now?

Answer 3

YES! thank you, I got A now ....can you help me solve any of the others?

Answer 4

sure... so you have omega f
and omega f = omega i + alpha *t

Answer 5

linear speed is V = r*omega

Answer 6

good on those two?

Answer 7

still a bit puzzled on "B"

Answer 8

so I'm still figuring out how to write it with the numbers given to me

Answer 9

\[\omega _{f} = \omega _{i} + \alpha t\]

Answer 10

it's just like the equation from kinematics... remember? Vf = Vi + at

Answer 11

Final velocity = to initial velocity + acceleration * Time?

Answer 12

here the quantities are all angular(except time of course) ... angular velocity, angular acceleration etc.

Answer 13

yep, that one.

Answer 14

okay now its coming to me a bit so final w is = to initial w + acceleration * time? "w" would stand for angular?

Answer 15

yes
\[\omega \]
is angular velocity
\[\alpha \]
is angular acceleration
\[\theta \]
is angular displacement

Answer 16

so here I'm gonna draw out my equation , which hopefully is write

Answer 17

right*

Answer 18

I got it now

Answer 19

so now I'm confused about c

Answer 20

what about it?

Answer 21

I know that the speed of the edge of the grinding wheel will be faster ,but really can't remember how

Answer 22

I just can't figure out the formula to use

Answer 23

V=r*omega

Answer 24

but in this case the number are .29m, 2705 rpm and 2 seconds

Answer 25

.29m is diameter, not radius.
we don't use rpm's we use rad.s/sec.

Answer 26

okay I think I should get this, might take me a while I'll message back if I need any help thanks

Answer 27

ok.

Answer 28

if the trampoline behaves like a spring with a spring stiffness constant k= 3 X 10^4 n/m , how far does the person depress it? I know the height = 3 and his speed as he lands on the trampoline is 10.38 m/s. What is the equation?