How do you solve these stroboscobe angular speed problems?

[shin777]

mass of color wheel is 133.4g and raidus is 12.7cm. The moment of inertia for disk is I = (1/2)mr^s. We used strobescope and color wheel and measured the time of how long it took till color wheel's yellow part looked stop. It took 2.1s for speed up and 1.5 second for speed down. Maximum value we could go with stroboscope was 525 flash/min.
second highest 475 flash/min
third highest 425 flash/min

these were what I got but I am not sure they are right. help please. :(

question 1) convert maximum angular speed of wheel to rev/s.
so the maximum speed we could go for strobescope and color wheel was 525 flash/min.

so 525/60 = 8.75 rev/s

question 2) convert maximum angular speed to wheel to rad/s

so 8.75*2pi = 53.4 rad/s

question 3)Find angular acceleration in rev/s^2 auuming it is constant.

σ = 8.75/2.1 = 4.17 rev/s^2

question 4)Repeat question 3 when the wheel was turned off.

σ = 8.75/1.5 = 5.83 rev/s^2

question 5)find the angle turned through during the acceleration in question 3, in rev.

Wa = (53.4 + 0)/2
26.7 = θ/2.1
θ = 56.7
56.7/2π = 9 rev

 

[barryj]

Looks good but the units on part 5 are wrong.

 

[shin777]

Looks good but the units on part 5 are wrong.

is it rev/s^2?

 

[barryj]

What are the units of an angle?

 

[shin777]

unit for angle is degree but the question is asking it in rev so I converted 56.7 degree to 9 rev by divide it by 2pi.

 

[barryj]

You can give the angle in degrees, revs, or radians. However, my comment is that rev/s in a velocity, not an angle.

 

[shin777]

oh.. so is it just rev then?

 

[barryj]

That is correct

 

[shin777]

That is correct

Thank you. Here are rest of questions. I think I got 6, 7 right but I don't understand from 8 to 11 at all. How do I do these? :(

question 6)Find I, the moment of inertia for the color wheel in kg m^2.
I = 1/2 mr^2
I = (1/2)(133.4)(12.7^2)
I = 10758/1000 = 10.758 kg m^2

question 7) find torque exerted by motor during the acceleration of the color of wheel in units of
N m.

T = Iσ
T = 10.758*5.83
T = 62.7 N m.

question 8)Multiply highest strobe setting which "stopped motion" by one. Multiply the second highest setting by two, and the third by three. Compare these products and then explain what is happening physically for the three highest settings which "stopped" the motion of the wheel. What is the wheel doing compared to the strobe flash rate for each of these three speeds?

525x1 = 525
475x2 950
425x3 = 1275

have no idea what to do next. :(

question9) State the true direction of motion of the disk.

*seems to go clock-wise to me but I don't think this is answer. :(

question 10)How will the disk appear to move relative to its true motion if the strobe is flashing slightly faster than the true rate of revolution, in the same direction or in the opposite direction?

*I have no idea on this one either. :(

question 11)Repeat question 10 for the strobe flashing slightly slower than the true rate of revolution.

*no idea on this one too. it seems like this is guess question? :(

 

[barryj]

I do not understand #8. For 10 and 11, think about this. If the strobe is at the same frequency as the color wheel then the yellow will be stopped. If the strobe is slightly faster than the wheel will the yellow be illuminated before or after the previous point and what will this look like?

Be careful on #7, consider the friction.

 

[shin777]

it didn't give friction value for #7 though..

as for 10, I think disk would appear to move slower and , 11 would make disk look move faster?

 

[barryj]

#7, no it didn't but can you determine the frictional force from the time it takes to slow down?
10 & 11, you did not answer the question.

 

[shin777]

can I get equation for #7 then?

10, I think it would appear to move slower in same direction and 11, it would appear to move faster in opposite direction. I am not really sure. :(

 

[barryj]

#7 Torque = (moment of inertia)(alpha)
What is alpha based on how long it takes to go from the full speed back down to zero.

On 10,11, Think about it again.

 

[shin777]

is it the other way around? I don't have strobescope and wheel in front of me so I can't guess. :(

as for #7, do i have to use 4.17 one instead of 5.83 as alpha?

 

[barryj]

When the wheel is slowing, the only torque is due to friction of the wheel so use the slowing down acceleration to find the frictional torque. When speeding up, the motor must have a torque that exceeds the frictional torque and enough in excess to cause an acceleration of 4.18.

On 10,11, visualize what is happening. Say you are looking at the second hand of a clock and the strobe flashes at exactly 60 second intervals. Let's also assume that it flashes when the second hand is at 12 o'clock. As long as the stays at exactly the same speed as the second hand, the hand will stay at 12 o'clock. Now what will happen if the strobe flashes just a little bit earlier, where whould you see the second hand. Would it be before or after 12 o'clock. Then on the next flash, where would you see the second hand. It will appear to move either in the same doirection as the second hand, cw, or in the opposite directio, ccw.

 

[shin777]

hmm.. i think it would make 10 look slower and ccw and 11 faster and cw.

so #7 is 10.758x4.17 = 44.9 N m.

 

[barryj]

OK on 10,11.
On #7 however. Recall that the NET torque = I X alpha .
What is the frictional torque when the wheel is slowing... T = I X 5.83.
Now when speeding up from the start, the NET torque must be the motor torque - the frictional torque and this must be I X acceleration when speeding up. I am trying not to give you the complete answer.

 

[shin777]

thx. what about question 9? do I need to solve question 8 to figure out question 9?

 

[barryj]

At this point, I still don't undersand #8.
As to 9, think about it like you did for10 and 11. If the strobe has initially stopped the motion, then if you increase the frequency of the strobe slightly, what would the disk look like depending on which way it was turning?