Question

Can anyone help me with this rotational physics problem

Answer 1

@phi @.Sam.

Answer 2

@zepdrix @DanJS

Answer 3

just read through it, ... you have anything yet?

Answer 4

It shows you how they got this relationship between the rotation speed and numbers of teeth on each gear

\[\large \frac{ \omega _{B} }{ \omega_{A} }=\frac{ N _{A} }{ N _{B} }\]

Answer 5

The prob says, the front has 4 times as many teeth as the rear, Call the front one A and the rear B, so

\[N _{A}=4*N _{B}\]

Answer 6

It also gives the angular speed of the front gear A
\[\omega _{A}=1.09\]

you can use those and calculate the angular speed of the rear gear B

Answer 7

im just sort of confused on what to do next exactly?

Answer 8

They give the rotation speed of the front gear, and the ratio of numbers of teeth for the two gears, you can use that to calculate the rear grear rotation speed,

\[\large \frac{ \omega _{B} }{ 1.09 }=\frac{ 4*N _{B} }{ N _{B} } =4~~~~~so~~~~~\omega
_{b}=4.36~~rev/s\]

the rear gear is rotating at 4.36 revolutions per second

Answer 9

We can assume that toe rear wheel is welded to the rear gear, so the rotation speed of the rear tire is the same as the rotation speed of the rear gear

Answer 10

ok then we just plug that in to the formula right

Answer 11

so basically 1.09/4.36=N(a)/N(b)

Answer 12

yeah, you can get the bike speed from how fast the rear wheel is turning..

rear wheel radius r=0.35 m
rear wheel turning at 4.36 rev/s

the tangential velocity is equal to the circle radius times its rotation speed

\[\large v=r*\omega _{B}\]

Answer 13

So the tangential velocity of the rear tire is

v = 0.35*4.36 = 1.526 m/s

that is how fast the bike is moving relative to the ground

Answer 14

but thats not the answer, right?

Answer 15

because its asking for the velocity of bike?

Answer 16

@DanJS

Answer 17

I have to submit the answer to this problem in the next 20 minutes @DanJS but I would also like to understand this concept.

Answer 18

@DanJS ?????

Answer 19

?

Answer 20

sorry, yeah i believe that is correct.

The rear gear spins at 4.36 rev/s, so in turn the rear wheel is spinning at the same 4.36 rev/s.
If the rear wheel is rotation at 4.36 rev/s and has a radius 0.35 m, you can get the tangential velocity for that rear wheel...

Answer 21

when i input 1.526m/s in the stem it is displaying as incoorect

Answer 22

oh i see what is wrong, for the tangential velocity

\[\large v=r*\omega \]
for the velocity to have meters per second, you should have the rotation speed in rad/s also, not rev/s

Answer 23

1 revolution per second is the same as 2pi radians per second

\[\large v=r*2\pi*\omega ~~~~~m/s\]

Answer 24

so 9.6m/s?

Answer 25

try that

v = 0.35 * 2pi*4.36
about 9.59 m/s,

yeah

Answer 26

alright sweet tysm

Answer 27

welcome