## Work Energy Method for Rotational Motion

1. The problem statement, all variables and given/known data
I don't how how to work out for the Gravitational Energy "H".

Info given is 1.5rev converting to 9.42rad/s
S=rδ
S=0.2x9.42
=1.884

*Should i multiply the axle radius or wheel?

3. The attempt at a solution
E1=K1+G1+S1
=1/2(8.5)202+(20)(9.81)(H)+0

Thanks
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 Quote by freshbox 1. The problem statement, all variables and given/known data I don't how how to work out for the Gravitational Energy "H". Info given is 1.5rev converting to 9.42rad/s
Check the units for the 9.42

 S=rδ S=0.2x9.42 =1.884 *Should i multiply the axle radius or wheel?
What are you trying to calculate above?

 3. The attempt at a solution E1=K1+G1+S1 =1/2(8.5)202+(20)(9.81)(H)+0
Your expression for the kinetic energy is not correct. How do you calculate the kinetic energy of a rotating object? The hanging mass also has kinetic energy at the initial time.

Since the mass is located at the "datum" at the initial time, you can consider the potential energy to be zero at that time.

 Sorry should be 9.42rad S=rδ S=0.2x9.42 =1.884 I am trying to calculate the distance for the height for gravitational energy How do you calculate the kinetic energy of a rotating object? K=Iω2 Since v=rω 2=0.1ω ω=20rad/s what do you mean by potential energy? do you mean kinetic? The equation should be: 1/2(8.5)(20)2+1/2(20)(2)2+G1+0 Rotational as "1/2(8.5)(20)2" Linear as "1/2(20)(2)2" G1-don't know how to form S1 as 0 because there is no spring

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## Work Energy Method for Rotational Motion

 Quote by freshbox Sorry should be 9.42rad S=rδ S=0.2x9.42 =1.884 I am trying to calculate the distance for the height for gravitational energy How do you calculate the kinetic energy of a rotating object? K=Iω2 Since v=rω 2=0.1ω ω=20rad/s
OK, that looks good except for a factor of 1/2 in K.
 The equation should be: 1/2(8.5)(20)2+1/2(20)(2)2+G1+0 Rotational as "1/2(8.5)(20)2" Linear as "1/2(20)(2)2" G1-don't know how to form
All right. I didn't know what the symbols S and G were standing for. As you say, there's definitely no spring potential energy. For the gravitational potential energy, note that the hanging mass is located initially at the "DATUM". That's the point where the gravitational potential energy is taken to be zero. So, G1 = 0.

 Did the box go down eventually? To my understanding yes because that question says in 1.5 rev" So that means the box travel 1.5 rev downwards. So i set my datum point at the 1.5 rev point, hence there is gravitational potential energy at the initial position which i am trying to calculate but i don't know whether to use the axle or the wheel radius.

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 Quote by freshbox Did the box go down eventually?

Yes. So, when you go to part (b) the box will be below the datum.

[EDIT: Did you use the correct radius for finding this distance?]

 Did the box go down eventually? To my understanding yes because that question says in 1.5 rev" So that means the box travel 1.5 rev downwards. So i set my datum point at the 1.5 rev point, hence there is gravitational potential energy at the initial position which i am trying to calculate but i don't know whether to use the axle or the wheel radius.
 Hmm after some thinking, do you mean i have to follow the question datum and i cannot set the datum myself unless the question never state the datum in the 1st place?

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 Quote by freshbox Did the box go down eventually? To my understanding yes because that question says in 1.5 rev" So that means the box travel 1.5 rev downwards. So i set my datum point at the 1.5 rev point, hence there is gravitational potential energy at the initial position which i am trying to calculate but i don't know whether to use the axle or the wheel radius.
You don't want to change the datum point. The answers you gave are for the datum shown in the picture. So, if you want to get the same answers, you need to keep the datum as shown.

The string tied to the box is wrapped around the pulley of radius 0.10 m. So, that's the radius you want to use to convert angle of rotation to linear distance traveled by the box.

 For Part B U1-2=T.δ+F.s =(P)(9.42)+(P)(0.942)
 Recognitions: Gold Member Homework Help For part (b) you just need to calculate the total energy at the final location. What is the kinetic energy there? What is the potential energy there? [Sorry, that's for part (c)]
 Recognitions: Gold Member Homework Help For part (b) you need to think about how to calculate the work done by a force. Do you know a formula for this?
 I don't know I am just trying to calculate the linear and rotatational energy. This is getting confusing Attached Thumbnails
 Recognitions: Gold Member Homework Help Confusion is normal. We'll get it. The questions in the problem are sort of guiding you through step by step. So, in part (a) you find the initial energy. In part (c) you will find the final energy. For part (b), you just need to find an expression for the work done by the force P. You have posted a couple of pages. One expresses the work in terms of force and distance, the other in terms of torque and angle. You can use either one of these expressions. They will yield the same answer.
 For Part B U1-2=T.δ+F.s =(P)(9.42)+(P)(0.942) Wrong answer I assume my linear work done is correct. But my δ is definitely wrong. *wait give me some time to work out

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 Quote by freshbox For Part B U1-2=T.δ+F.s =(P)(9.42)+(P)(0.942)
Use either force times distance or torque times angle. If you decide to use force times distance, the distance must be linear distance in meters. If, instead, you decide to use torque times angle to get the work, then you will need to express the torque in terms of P.

 "Use either force times distance or torque times angle." I thought there is linear and rotational? The parcel is moving downwards (linear) and the wheel is rotating (rotational) ???
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