Solving for Angular Velocity: Steel Block on Rotating Table

[helpppmeee]

Homework Statement

A 500 g steel block rotates on a steel table while attached to a 2.0-m-long massless rod. Compressed air fed through the rod is ejected from a Inozzle on the back of the block, exerting a thrust force of 3.5 N. The nozzle is 70 degrees from the radial line, so that the net force created by the nozzle points 20 degrees, from the tangential velocity, toward the inside of the cirlce . The block starts from rest. What is the angular velocity after 10 rev?

Homework Equations

The Attempt at a Solution

so i found the acceleration along the tangential velocity, used the equation; xF = xI + vI*t + 1/2(a)t^2, where xF and xI are the final and initial displacements, respectively (vI[initial velocity] is clearly zero). xF = 20*pie*radius and i found the time it takes to make 10rev. I then took the total radians in 10 rev (20*pie) and divided it by the total time. i got 10.something rad/s where the answer is actually 6.63rad/s. I also found the centripetal force after those 10 revs and tried to do the same thing, again got the same answer. helpp pleeeeasee!

 

[p21bass]

The nozzle is 70 what from the radial line?

 

[helpppmeee]

70 degrees

 

[helpppmeee]

bumpity bump bump (mid term in 2 hours. Desperate for help!) i made the question somewhat clearer

 

[p21bass]

OK - so I did the problem out, using the method you tried, and got the 6.63 rad/s answer.

Can you show all your work, so I can see where you went wrong?

My initial guess is that you: didn't include the frictional force of the steel on steel and/or neglected the angle of F. But I'm not sure.

 

[helpppmeee]

i used t = sqrt(126/3.3) [3.3 being half the acceleration] then w = (20*pie)/t

 

[p21bass]

I have no idea how you came up with those numbers, as 3.3 (what?) is nowhere near either the angular or tangential acceleration.

And 126 (again, what?) doesn't correspond to anything I can think of, either.

You NEED to show all your work, as you've apparently made at least one rather large error.

 

[helpppmeee]

sorry about that i figured u knew what i was doing with the xF = xI + vI*t + 1/2(a*t^2)... the vI is zero, so i substituted 126m (the circumference of 10 revs) into xF. The force i found was Ft[force along tangential velocity]=3.5cos(20)... I did not factor in a frictional force after this (my error). I then took that force and got a tangential acceleration of 6.6m/s^2. I then substituted this into the xF = xI + vI*t + 1/2(a*t^2) equation and solved for time. After finding the time it takes to make 10 revs, i took the total angle in rads and divided it by the time it took to make those revs w = (2*pie*10)/t. I tried it once more with the friction factored in pointing in the opposite direction of tangential acceleration and i still didnt get the correct answer (6.6rad/s)

 

[p21bass]

OK, so I see where you got 3.3 from before. But, that's not your acceleration - that's the amount of Force in the tangential direction. You'd need to plug that into the sum(F) = ma equation, to get your acceleration. That seems like it's where you went wrong. And, yes, you'll have to use the coefficient of kinetic friction for steel on steel. If it's given (should be, or at least in a table in your text), use that. Otherwise, it's generally accepted as 0.6.

 

[helpppmeee]

t^2 = xF/((1/2)*a) where a = 6.6... Factoring in friction i get a net force of 0.36N in the tangential Velocity direction. i solve for t (sqrt(xF/((1/2)*0.72)) which is 18.7s. 18.7*0.72=velocity, which is 13.5 m/s. (Fnet)r = (0.5*13.5^2)/2 = 91N... 91N/(0.5*2) = w^2... sqrt that and i get 6.74... what am i doing wrong?

 

[p21bass]

Factoring in friction i get a net force of 0.36N in the tangential Velocity direction. i solve for t (sqrt(xF/((1/2)*0.72)) which is 18.7s. 18.7*0.72=velocity, which is 13.5 m/s.

That much is correct. There's an easier way to do it, where you can obtain the velocity without calculating t (v^2 = (v_0)^2 + 2a(xF - xI)). But your method works, too.

But I'm not sure why you tried to do this... Yes, it uses centripetal acceleration, but there's a much easier way to do it, at this point.

(Fnet)r = (0.5*13.5^2)/2 = 91N... 91N/(0.5*2) = w^2... sqrt that and i get 6.74... what am i doing wrong?

If you know the linear velocity and the radius, you can simply plug those numbers into the equation w=v/r. Using your answer for v, you get a slightly different value for w, but that's just because of rounding intermediate answers. Really, what you did here works - and I think it's just a matter of rounding, as 6.74 ~ 6.63.
I'll post how I did it, next.

 

[p21bass]

Given
m = 500g = 0.5kg
r = 2.0m
n = 10 rev
F = 3.5N
Θ = 70 degrees
mat'l = steel on steel => μ = 0.6
v₀ = 0
s₀ = 0

Objective
ω


Procedure

F_N = W = m*g = 0.5kg*9.81m/s² = 4.905N
F_applied = F*sinΘ = 3.5N * sin(70°) = 3.29N
F_friction = F_N * μ = 4905N * 0.6 = 2.94N
s = n * 2π * r = 10rev * 2πrad/rev * 2.0m = 40πm

ΣF = m*a => (F_applied - F_friction) = m*a
Rearranging => a = (F_applied - F_friction) / m
a = (3.29N - 2.94N) / 0.5kg = 0.7m/s²

v² = v₀² + 2*a*(s - s₀)
v² = 0² + 2 * 0.7m/s² * (40πm) = 175.8276m²/s²
v = sqrt(175.8276m²/s²) = 13.26m/s

ω = v / r = (13.26m/s)/(2.0m) = 6.63 rad/s