What are the torque and power calculations for a belt drive system?

[The Futur]

Homework Statement

(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.

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 I am on the right track or I am totally off

 

[Redbelly98]

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 don't 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 don't 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,

r_A ω_A = v_A = v_B = r_B ω_B​

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]

r_A ω_A = v_A = v_B = r_B ω_B​

In other words,

r_A ω_A = r_B ω_B​

(ii) asks for ω_B. You know what r_A, ω_A, and r_B are. Solve the equation for ω_B.

 

[The Futur]

In other words,

r_A ω_A = r_B ω_B​

(ii) asks for ω_B. You know what r_A, ω_A, and r_B are. Solve the equation for ω_B.

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 weird answer

 

[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 weird answer

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.