Im starting to understand.

[The Futur]

1. The problem statement, all variables and given/known data

http://i138.photobucket.com/albums/q275/XxmassxX/fg6.png

(b) Fig. Q.6 shows a belt drive system where pulley A (diameter 0.5 m) is transmitting power to pulley B (diameter 0.25 m), mounted on a parallel rotating shaft. The tight and slack-side tensions of the belt are 500 N and 200 N respectively, and pulley A rotates at 300 rev/min.

Determine
(i) The linear speed of the belt (in m/s))
(ii) The rotational speed of pulley B (in rev/min)
(iii) The torque on pulley B.
(iv) The power transmitted to pulley B.





3. The attempt at a solution

(i)
To find linear speed in (m/s)

v= RxW
V=0.5x(300x(2pi/60))
v=0.5x31.4159
v=15.7079m/s

(II)
the rotational speed of pully B

if
0.5m = 300rev/min
therefore
300/0.5 = 600
600x0.25= 150rev/min

(iii)
To find torque

Tb=(f1-f2)Rb
Tb=(500-200)0.125
Tb=(300)0.125
Tb=37.5Nm

(iv)
To find power

Pb=TbxWb
Pb=37.5x(150x(2pi/60))
Pb=589.04watts

i did my best, can some one tell me if im on the right track or im totaly off

[Redbelly98]

3. The attempt at a solution

(i)
To find linear speed in (m/s)

v= RxW
V=0.5x(300x(2pi/60))
v=0.5x31.4159
v=15.7079m/s

You have the right idea. However, the radius of pulley A is not 0.5m.


(II)
the rotational speed of pully B

if
0.5m = 300rev/min

That statement makes no sense.


therefore
300/0.5 = 600
600x0.25= 150rev/min

Not quite. Hint: both pulleys have the same linear speed.


(iii)
To find torque

Tb=(f1-f2)Rb
Tb=(500-200)0.125
Tb=(300)0.125
Tb=37.5Nm

Yes, that's right.


(iv)
To find power

Pb=TbxWb
Pb=37.5x(150x(2pi/60))
Pb=589.04watts

You have the right idea, but will need to use the correct value of Wb to get the correct answer here.

[The Futur]

(i)
To find linear speed in (m/s)

v= RxW
V=0.25x(300x(2pi/60))
v=0.25x31.4159
v=7.85m/s
is that corect now?

i dont get you on the rotational speed of pully B.

if i get the rotational speed of the pully B, i will be able to have the correct value for Wb yeah?

[djeitnstine]

\omega_b = 300 \frac{rev}{min} \left( 2 \frac{\pi}{60} \right) not \omega_b =150 \frac{rev}{min} \left( 2 \frac{\pi}{60} \right) (in your power equation.)

[The Futur]

\omega_b = 300 \frac{rev}{min} \left( 2 \frac{\pi}{60} \right) not \omega_b =150 \frac{rev}{min} \left( 2 \frac{\pi}{60} \right) (in your power equation.)


yeah i do know it's 300rev/min, but how did u find the 300rev/min?

so to answer the question.

Q. The rotational speed of pully B
A. Since pully A as a rotational speed of 300rev/min
and both pully have the same linear speed, so pully b as a rotational speed of 300rev/min?

[Redbelly98]

i dont get you on the rotational speed of pully B.

Okay. You know that the linear speed v is

v = r ω

And since the linear speed (= the belt speed) is the same for both pulleys,

rA ωA = vA = vB = rB ωB

Also, B will have a different rotational speed (not 300 rpm) than A.

if i get the rotational speed of the pully B, i will be able to have the correct value for Wb yeah?
Yes.

[The Futur]

im confused!!

[Redbelly98]

rA ωA = vA = vB = rB ωB


In other words,

rA ωA = rB ωB

(ii) asks for ωB. You know what rA, ωA, and rB are. Solve the equation for ωB.

[The Futur]

In other words,

rA ωA = rB ωB

(ii) asks for ωB. You know what rA, ωA, and rB are. Solve the equation for ωB.

So
Va=RaxWb = Vb=RbxWb

Since Va=7.85m/s

Vb=RbxWb
7.85=0.125xWb
0.125Wb=7.85
Wb=7.85/0.125=62.8rev/min
is that correct?

[Redbelly98]

Almost, except for the units. It's 62.8 rad/sec.

[The Futur]

Almost, except for the units. It's 62.8 rad/sec.

how do i convert taht to rev/min.

i tryed

62.8/(2pi/60)
and i got some wierd answer:yuck:

[Redbelly98]

how do i convert taht to rev/min.

i tryed

62.8/(2pi/60)

Good, that is the correct way to convert from rad/s into rev/min.


... and i got some wierd answer:yuck:
I get an entirely reasonable, and correct, answer when I do the calculation.

p.s.
If you had posted the answer you actually got, I could now be posting back about whether you did the calculation correctly.