Impulse Momentum Method for Rotational

In summary, the calculations for torque and angular velocity are incorrect because the radius of the wheel is not taken into account.
  • #1
freshbox
290
0

Homework Statement


I have abit of problem understanding this 2 question, I would appreciate if someone can guide me along.

For diagram 1 (Left)
I interpret the condition as: At 0sec, initial velocity is 0, At 5sec final velocity is unknown.


For diagram 2 (Right)
I interpret the condition as: At 0 sec, initial velocity is 0, At 3sec final velocity is 14.5m/s


Please let me know if I get any of the condition wrong, thank you.
 

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  • #2
I think you understand the conditions correctly.
 
  • #3
o_0 hmm strange then how come i cannot get the answer...
 
  • #4
What goes wrong?
 
  • #5
For Diagram 1 part A

M1=mv
=0

M2=mv
=(20)(V2)
=20V2

M1+I1-2=M2
981-5TA=20V2
5TA=981-20V2
TA=196.2-4V2

Ans is TA=196.2-4V

4V2 and 4V are both different meaning right?
 
  • #6
Why are you ignoring the tangential force?
 
  • #7
You mean the 75N force?

What has it got to do with part A?
 
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  • #8
You have already solved part A as far as I can tell.
 
  • #9
My Answer is: TA=196.2-4V2
Question Ans is : TA=196.2-4V

Is there any difference between V2 and V? one is initial one is final.
 
  • #10
The initial velocity is zero per the conditions ("released at t = 0"). So v can only be the final.
 
  • #11
How about the subscript? Don't have to write?
 
  • #12
Up to you. You can say "let's denote the velocity at t = 5 s as V" or in any other way. That does not change the underlying physics. You just need to know what denotes what.
 
  • #13
Part B

M1=Iω
=0

M2=Iω
=(2.5)(V/0.15)
=16.67V

I1-2=Torque x Time
=[-22.5+Ta(0.15)]5
=34.65-3V

M1+I1-2=M2
34.65-3V=16.67V
34.65=19.67V
V=1.76

V=rω
ω=1.76/0.15
ω=11.74 - Ans Wrong
 
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  • #14
freshbox said:
ω=11.74 - Ans Wrong

Hmm. I have used a somewhat different method, but got the same result. Let's see if somebody else here can point out our mistake.
 
  • #15
freshbox said:
M2=Iω
=(2.5)(V/0.15)
=16.67V
OK.

I1-2=Torque x Time
=[-22.5+Ta(0.15)]5
=34.65-3V
Where did you get the 22.5 from? Redo that calculation.
 
  • #16
Hi Doc Al, thanks for helping.

I got my -22.5 from the tangential force Ft=75N I used the 75 times 0.3 = 22.5 Because Tension x Radius = Torque


*Ok i think i got it, it's 300mm, 150mm diameters
 
  • #17
freshbox said:
*Ok i think i got it, it's 300mm, 150mm diameters
There you go. :wink:
 
  • #18
M1=Iω
=0

M2=Iω
=(2.5)(V/0.15)
=16.67V

I1-2= Torque x Time
=(TaX0.075-75X0.15)5
=17.325-1.5V

M1+I1-2=M2
17.325-1.5v=16.67v
17.325=18.17v
v=0.95

v=rω
0.95=0.15ω
ω=6.3 wrong again :cry:
 
  • #19
freshbox said:
M2=Iω
=(2.5)(V/0.15)
=16.67V
Same problem with radius vs diameter. (Sorry I didn't spot it before.)
 
  • #20
how come i need to take the radius of the axle and not the wheel?
 
  • #21
freshbox said:
how come i need to take the radius of the axle and not the wheel?
You are relating ω to the speed V, which is the tangential speed of the axle edge, not the wheel edge.
 
  • #22
Can I say the "v" that I am finding is actually the velocity of the axle. After getting the velocity of the axle, I divide by 0.075 (axle radius) is because in a compound pulley, Angular velocity are the same. I am just using the axle velocity to get ω.
 
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  • #23
freshbox said:
Can I say the "v" that I am finding is actually the velocity of the axle. After getting the velocity of the axle, I divide by 0.075 (axle radius) is because in a compound pulley, Angular velocity are the same. I am just using the axle velocity to get ω.
Yes. V is the velocity of the falling mass, and since its cord is wrapped around the axle, V is also the tangential velocity of the axle.

The axle and wheel are attached, so they have a common ω.
 
  • #24
Ok..I'm sorry can you help me take a look at this question part C

M1=Iω
=(3.5)(12.5)
=43.75

M2
=Iω
=3.5ω

I1-2=Torque X Time
=[(-TA)(0.2)+(TB)(0.5)]5
=-32ω+3783.5

M1+I1-2=M2
43.75-32ω+3783.5=3.5ω
3827.25=35.5ω
ω=107.80 - Wrong Answer
 

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  • #25
freshbox said:
M1=Iω
=(3.5)(12.5)
=43.75

M2
=Iω
=3.5ω

I1-2=Torque X Time
=[(-TA)(0.2)+(TB)(0.5)]5
=-32ω+3783.5
Double check that last calculation.
 
  • #26
=[(-TA)(0.2)+(TB)(0.5)]5

=[(-2ω+465.5)0.2+(1327.2-12ω)0.5]5

=(-0.4ω+93.1+663.6-6ω)5

=(-6.4ω+756.7)5

=-32ω+3783.5

I'm abit blind hehe can you tell me where is wrong? :redface:
 
  • #27
freshbox said:
=[(-TA)(0.2)+(TB)(0.5)]5

=[(-2ω+465.5)0.2+(1327.2-12ω)0.5]5
Looks to me like you left out a minus sign.
 
  • #28
fail -.- what a mess...hehe anyway thanks for your time and help (voko too), thanks :)
 
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  • #29
I can't believe I mistook the diameters for the radii, but I did. Not just once, but multiple times when I cross-checked my results via at least three different methods. Oh well. Thanks Doc Al for pointing that out.
 

1. What is the impulse momentum method for rotational motion?

The impulse momentum method for rotational motion is a technique used to analyze the motion of objects that are rotating about a fixed axis. It is based on the laws of conservation of momentum and angular momentum, and involves calculating the change in angular momentum of an object due to an impulse or external force.

2. How is the impulse momentum method different from the linear impulse momentum method?

The impulse momentum method for rotational motion is similar to the linear impulse momentum method in that it involves calculating the change in momentum of an object. However, the rotational method takes into account the object's moment of inertia and angular velocity, whereas the linear method only considers mass and linear velocity.

3. What is the equation for calculating angular impulse in the impulse momentum method for rotational motion?

The equation for calculating angular impulse is given by J = ΔL = IΔω, where J is the angular impulse, ΔL is the change in angular momentum, I is the moment of inertia, and Δω is the change in angular velocity.

4. How is the impulse momentum method used to solve rotational motion problems?

To solve a rotational motion problem using the impulse momentum method, you first need to determine the initial and final angular momentum of the object. Then, you can use the equation J = ΔL = IΔω to calculate the angular impulse. Finally, you can use this impulse to find the final angular velocity of the object.

5. Can the impulse momentum method be used for all types of rotational motion?

Yes, the impulse momentum method can be used for all types of rotational motion, as long as the object is rotating about a fixed axis. It can be applied to problems involving both constant and variable angular velocity, as well as problems with multiple objects and collisions.

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