How do I solve for rotational dynamics using Newton's second law?

In summary: I used the relationship between tangential acceleration and alpha at=Rα and subbed in α resulting in T1=(2/5)Mat 4.then I summed the torques for the disk and solved...5.then I subtracted T1 from T2 and solved for at6.at=(m2g)/((2/5)Mat+m2-(1/2)m1)In summary, Homework Equations state that the moment of inertia of a sphere is 2mR^2/5, and the moment of inertia of a disk ismr^2/2. V(center of mass) is V(tangential) - because of smooth
  • #1
Dusty912
149
1

Homework Statement


I attached the stated questions. Mainly need help with question 2. See below for my answer to question 1.

Homework Equations


ΣT=Iα
ΣF=ma
moment of inertia of sphere: I=(2mR^2)/5
moment of inertia of disk: I=(mr^2)/2
v=rω
V(center of mas)=V(tangential) -because of smooth rolling
ΔΘ=ωt+(1/2)αt^2
ω(final)=ω(initial)+αt
ω^2(final)=ω^2(initial) +2αΔΘ

The Attempt at a Solution


1. C) 4F1 -I figure the lever arm is 1/4 as big for the smaller disk so it would require a force 4 times as big to counteract the force applied to the larger disk.

2. a) I know how to draw the FBD for this.
B) what I need the most help with. It has been a while since we have done this section and I had trouble with it the first time. But I am assuming we use the analogy for Newton's second law ΣT=Iα and the sum the torques. But do I do this for the sphere and the the disk together? not really sure. My guess is that it would look something like this. T1=the tension in the rope tangential to the sphere and disk.
T2=the tension in the rope tangential to the disk and connected to the block.
So summing the torques for the sphere: T1R=(2/5)mR^2α -the radius would be of the sphere of course

Summing of the torques for the disk: T2r=(1/2)mr^2α

and summing of the forces for the block would be T2-mg=ma (the m's would be the mass of the block)
from here I'm stuck. Sorry I couldn't show more work but like I said I really need help with this one.

Part (c) should be pretty easy once I know the acceleration of the black from part B
 

Attachments

  • Quiz Rotational Dynamics TAKE HOME FA15.pdf
    90.5 KB · Views: 330
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  • #2
Basically, there are two tensions in the string.
One tension accelerates the sphere.
The other tension is between the disk and the mass.
The difference of these tensions accelerates the disk.
 
  • #3
Okay that somewhat helped. But I'm not really sure what I should do with that information. Could you elaborate? or anyone else care to add on?
 
  • #4
okay so this is what I have so far: I summed the forces for the block resulting in T2=m2g-m2a
then I summed the torques for the sphere resulting in: T1R=(2/5)MR^2α then I used the relationship between tangential acceleration and alpha at=Rα and subbed in α resulting in T1=(2/5)Mat then I summed the torques for the disk resulting in : T1r-T2r=Iα then subbing in α and the moment of inertia for a disk and canceling all of the radii yields T1-T2=(1/2)m1at subbing in T1 and T2 and solving for at yields
at=(m2g)/((2/5)Mat+m2-(1/2)m1)
which gave me 2.58m/s

Am I in the ballpark?

oh and by the way I'm now using the variables from the attached document, sorry for the confusion
 
  • #5
Just to clarify, those two uses of alpha are for different angular accelerations, yes?
 
  • #6
haruspex said:
Just to clarify, those two uses of alpha are for different angular accelerations, yes?
yes they are, sorry knew I should have clarified that
 
  • #7
Dusty912 said:
yes they are, sorry knew I should have clarified that
Ok.
Your equation for at in post #4 doesn't look right. Does it make sense that increasing the mass of the disk would increase the acceleration?
Even with the equation as you have it, your final numerical answer looks about ten times too large.
 
  • #8
okay so are you saying I did something wrong in my algebra?
 
  • #9
haruspex said:
Ok.
Your equation for at in post #4 doesn't look right. Does it make sense that increasing the mass of the disk would increase the acceleration?
Even with the equation as you have it, your final numerical answer looks about ten times too large.

wouldn't increasing the mass of the disk make the acceleration smaller? just for clarification m2 is the block and m1 is the black
 
  • #10
Dusty912 said:
wouldn't increasing the mass of the disk make the acceleration smaller? just for clarification m2 is the block and m1 is the black
Quite so, but your equation says otherwise. Check the signs. If you cannot spot the error, post all your steps.
(The black?)
 
  • #11
haruspex said:
Quite so, but your equation says otherwise. Check the signs. If you cannot spot the error, post all your steps.
(The black?)
sorry m1 is the disk, and I am looking at it now. So I'm on the right track then?
 
  • #12
Dusty912 said:
sorry m1 is the disk, and I am looking at it now. So I'm on the right track then?
Yes, there is just the one error in the equation, and it looks like you had an arithmetic error later too.
 
  • #13
haruspex said:
Yes, there is just the one error in the equation, and it looks like you had an arithmetic error later too.
Okay so I'm still getting the same answer here are my steps:
1.I summed the forces for the block and solved for T2 T2=m2g-m2a
2.then I summed the torques for the sphere and solved for T1 and subbed in a=Rα for α. The R's cancel and yield T1=(2/5)Ma
3.then I sum the torques for the disk and sub in a=rα for α. That yields: T1-T2=(1/2)m1a
4. then I substitute T1 and T2 into the disk torque summation which yields: (2/5) Mat-m2g +m2at=(1/2)m1at
5. I move the terms with at onto one side and factor out at yielding: at((2/5)M +m2-(1/2)m1)=m2g
6. then I solve for at yielding at=(m2g)/((2/5)M+m2-(1/2)m1)
 
  • #14
Dusty912 said:
3.then I sum the torques for the disk and sub in a=rα for α. That yields: T1-T2=(1/2)m1a
Which ways do T1 and T2 act? Which way does the disk accelerate?
 
  • #15
haruspex said:
Which ways do T1 and T2 act? Which way does the disk accelerate?

oh I see so:
at=(m2g)/((1/2)m1 +m2 +(2/5)M)

?
 
  • #16
Dusty912 said:
oh I see so:
at=(m2g)/((1/2)m1 +m2 +(2/5)M)

?
Yes.
 
  • #17
okay thank you and as far as question #1 goes would C) 4F1 be correct?
 
  • #18
haruspex said:
Yes.
okay thank you and as far as question #1 goes would C) 4F1 be correct?
 
  • #19
Dusty912 said:
okay thank you and as far as question #1 goes would C) 4F1 be correct?
Yes.
 
  • #20
haruspex said:
Yes.
So I already turned in my quiz, but I was a little curious about the radii canceling out. Since that was information given to me, I figured Id have to use them.
 
  • #21
Dusty912 said:
So I already turned in my quiz, but I was a little curious about the radii canceling out. Since that was information given to me, I figured Id have to use them.
In the real world, there may be a lot of available information that turns out to be redundant in solving the problem at hand. So I endorse the provision of surplus information in questions. Were that done more commonly, it would not have bothered you that information provided was not used. Note also you were admonished to work purely algebraically until the final step. Had you plugged in numbers along the way, you might never have noticed that the radii were not relevant.
 

1. What is rotational dynamics?

Rotational dynamics is a branch of physics that studies the motion of objects that rotate around a fixed axis. It involves analyzing the forces and torques acting on an object to determine its rotational motion.

2. How is rotational dynamics different from linear dynamics?

In linear dynamics, the motion of an object is described in terms of its position, velocity, and acceleration along a straight line. In rotational dynamics, the motion is described in terms of angular displacement, angular velocity, and angular acceleration around a fixed axis.

3. What is the relationship between rotational dynamics and torque?

Torque is a measure of the force that causes an object to rotate around an axis. In rotational dynamics, torque is used to calculate the angular acceleration of an object, which in turn determines its rotational motion.

4. How does rotational inertia affect an object's motion?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to changes in its rotational motion. Objects with a higher rotational inertia will require more torque to achieve the same angular acceleration as objects with a lower rotational inertia.

5. What are some real-life examples of rotational dynamics?

Rotational dynamics can be observed in various everyday activities, such as swinging on a playground swing, spinning a top, or throwing a frisbee. It also plays a crucial role in the functioning of many machines, such as engines, turbines, and gears.

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