How to solve for the moment of inertia and angular acceleration?

  • #1
tellas
A girl is climbing up a set of stairs and has her femur horizontal. Her thigh weighs 12 pounds, acting at a center of mass 15 cm from her hip joint. She places 50 pounds of weight on the lower portion of that leg, vertically through her knee. Her gluteus maximus is applying a force directed 30 degrees below the horizontal, at a point 8 cm from her hip joint. An unknown force is applied at her hip.

1. How much force is her gluteus maximus exerting on her femur?
2. WHat are the magnitude and direction of the unknown force applied to her hip?
3. Approximate her thigh as a thin rod, with a mass of 12 pounds. What is her thigh's moment of inertia about either end?
4. If she increases the force from her gluteus maximus to 500 pounds, with everything else being the same, what angular acceleration will her thigh experience about her hip?

67ff2e1e9c004d318b5ff332f0a33747_A.jpg


I believe relevant equations would be:
angular acceleration= T/I
I=1/3Ml^2


For number 3 I tried to solve for moment of inertia by doing I= 1/3(5.4kg)(.4^2)
I am really confused on how I would solve the rest of the problems.
 

Attachments

Answers and Replies

  • #2
CWatters
Science Advisor
Homework Helper
Gold Member
10,530
2,297
You need to show a bit more effort to meet forum rules.

For 1). Try assuming there is no angular acceleration.
 
  • #3
tellas
You need to show a bit more effort to meet forum rules.

For 1). Try assuming there is no angular acceleration.
If I assumed there was no angular acceleration would I use F=ma?
 
  • #4
CWatters
Science Advisor
Homework Helper
Gold Member
10,530
2,297
If I assumed there was no angular acceleration would I use F=ma?
If there is no acceleration then it becomes a statics problem. What does that mean for the net force (F) or the net torque (T)?
 
  • #5
tellas
If there is no acceleration then it becomes a statics problem. What does that mean for the net force (F) or the net torque (T)?
The net force or torque would be equal to zero. So the sum of all the forces on the femur would be equal to zero?
 
  • #6
jbriggs444
Science Advisor
Homework Helper
9,487
4,172
The net force or torque would be equal to zero. So the sum of all the forces on the femur would be equal to zero?
Net force AND net torque.

F=ma. If the femur is not accelerating, the sum of the forces must be zero.

##\tau=I\alpha## (torque = moment of inertia times angular acceleration). If the femur has zero angular acceleration, the sum of the torques must also be zero.
 
  • #7
CWatters
Science Advisor
Homework Helper
Gold Member
10,530
2,297
The net force or torque would be equal to zero. So the sum of all the forces on the femur would be equal to zero?
So can you write sum equations now :-)
 
  • #8
tellas
So can you write sum equations now :-)
I am confused on how the sum of all the forces on the femur would equal zero since all of the given forces are positive
 
  • #9
jbriggs444
Science Advisor
Homework Helper
9,487
4,172
all of the given forces are positive
All of the given forces are vectors.
 
  • #10
CWatters
Science Advisor
Homework Helper
Gold Member
10,530
2,297
It sounds like you need to revise how to do statics problems. If you define "up" as positive then two of the forces have positive components in that direction and two have negative components in that direction.

Personally for the first question I would start by summing torques. Do you know how to pick a suitable point to sum the torques about?
 
  • #11
tellas
It sounds like you need to revise how to do statics problems. If you define "up" as positive then two of the forces have positive components in that direction and two have negative components in that direction.

Personally for the first question I would start by summing torques. Do you know how to pick a suitable point to sum the torques about?
Is the femur the suitable point to sum the torques? From what I understand, there are two forces in the x (hip and gluteus) direction and four forces in the y direction (hip, gluteus, thigh, and tibia) . The torque for the thigh would be (53.38)(.15)= 8.007 and the torque of the tibia would be (222.4)(.4)= 89.76. If I assume up is positive then tibia's torque is positive and thigh's torque is negative. I do not know how I can solve for the gluteus without knowing the hip's force.
 
  • #12
jbriggs444
Science Advisor
Homework Helper
9,487
4,172
Is the femur the suitable point to sum the torques?
The femur is not a point at all.
I do not know how I can solve for the gluteus without knowing the hip's force.
If you have one or more unknown forces, consider summing the torques about the point at which one of them acts.
 
  • #13
CWatters
Science Advisor
Homework Helper
Gold Member
10,530
2,297
Is the femur the suitable point to sum the torques? From what I understand, there are two forces in the x (hip and gluteus) direction and four forces in the y direction (hip, gluteus, thigh, and tibia) The torque for the thigh would be (53.38)(.15)= 8.007 and the torque of the tibia would be (222.4)(.4)= 89.76. If I assume up is positive then tibia's torque is positive and thigh's torque is negative. I do not know how I can solve for the gluteus without knowing the hip's force.
It's a statics problem so the net torque about any point is zero. As jbriggs suggests, if you sum the torques about the point where an unknown force acts then it disappears from the equation (because torque = force * distance and the distance part would be zero).

Since you are working with torques you should define clockwise/anticlockwise as positive rather than up/down.
 
  • #14
tellas
It's a statics problem so the net torque about any point is zero. As jbriggs suggests, if you sum the torques about the point where an unknown force acts then it disappears from the equation (because torque = force * distance and the distance part would be zero).
Since you are working with torques you should define clockwise/anticlockwise as positive rather than up/down.
Ok I think I am starting to get this. So if I choose the hip as my axis of rotation, then that would make the hip's torque 0. Then -Tgluteus - Tthigh +Ttibia = 0. Then I enter in the force and distances to get the torques so I would have - Fglutues x Xgluteus - Fthigh x Xthigh + Ttibia x Xtibia= 0. So then I would move it to solve for the force of gluteus by making it -Fgluteus= ((+Fthigh x Xthigh) - (Ftibia x Xtibia)) / Xgluteus. -Fgluteus = ((53.4 N x .15 m) - (222.4 N x .4 m))/ .08. -Fgluteus= -845 N. Is this correct or would I have to make the gluteus the axis of rotation?
 
  • #15
jbriggs444
Science Advisor
Homework Helper
9,487
4,172
Then I enter in the force and distances to get the torques so I would have - Fglutues x Xgluteus - Fthigh x Xthigh + Ttibia x Xtibia= 0
What about the angle of application for the forces?
 
  • #16
CWatters
Science Advisor
Homework Helper
Gold Member
10,530
2,297
Ok I think I am starting to get this. So if I choose the hip as my axis of rotation, then that would make the hip's torque 0. Then -Tgluteus - Tthigh +Ttibia = 0.
Yes that's the best approach.

Then I enter in the force and distances to get the torques so I would have - Fglutues x Xgluteus - Fthigh x Xthigh + Ttibia x Xtibia= 0. So then I would move it to solve for the force of gluteus by making it -Fgluteus= ((+Fthigh x Xthigh) - (Ftibia x Xtibia)) / Xgluteus. -Fgluteus = ((53.4 N x .15 m) - (222.4 N x .4 m))/ .08. -Fgluteus= -845 N. Is this correct or would I have to make the gluteus the axis of rotation?
The lever arm/distance you want is the perpendicular distance... .

 
  • Like
Likes jbriggs444
  • #17
tellas
So using that video I found the perpendicular lever arm of the gluteus force by using sin(30)(8)= 4cm. I then get (-(12lbs x 15cm)+(50lbs x 40cm))/4 and get a force of 455 pounds. Is this better?
 
  • #18
CWatters
Science Advisor
Homework Helper
Gold Member
10,530
2,297
Looks right to me.
 
  • #19
tellas
For question number 2, I set all of the forces equal to zero then moved the hip force by itself. Fgluteus+Fthigh-Ftibia= Fhip. 455 + 12 - 50 = +417lbs.
For question number 3, I used the equation I=(1/3)(mthigh)(lthigh)^2. I am not sure if I have to convert the units to kg and m, but if I do convert, I get (1/3)(5.4kg)(.4m)^2= .288 kg x m^2.
Am I doing these two questions correctly?
 
  • #20
CWatters
Science Advisor
Homework Helper
Gold Member
10,530
2,297
For question number 2, I set all of the forces equal to zero then moved the hip force by itself. Fgluteus+Fthigh-Ftibia= Fhip. 455 + 12 - 50 = +417lbs.
That's not correct. You don't know what direction the force at the hip acts at. You have two unknowns (magnitude and direction) so you need to write two equations... a sum of the vertical components to zero and a sum of the horizontal components to zero. Solve to give the two unknowns.
 
  • #21
tellas
That's not correct. You don't know what direction the force at the hip acts at. You have two unknowns (magnitude and direction) so you need to write two equations... a sum of the vertical components to zero and a sum of the horizontal components to zero. Solve to give the two unknowns.[/QUOTE

So the answer that I gave, is that correct for the vertical component? I'm a little confused on how I find the force of the gluteus vertically and horizontally. Would I use cos(30)= x/455lbs= 394 and sin(30)= x/455lbs= 227.5? If so then I would have 227.5+12-50= 189.5lbs for the vertical hip force. For the horizontal component, 394lbs=-Fhip. So Fhip=-394lbs.
 
  • #22
CWatters
Science Advisor
Homework Helper
Gold Member
10,530
2,297
I'm a little confused on how I find the force of the gluteus vertically and horizontally. Would I use cos(30)= x/455lbs= 394 and sin(30)= x/455lbs= 227.5?
Yes that's the right approach if you mean the x component of Fglutes is 394lbs and the y component is 227.5lbs.

Edit: There is nothing wrong with defining down and left as +ve but its more usual to define up and right as +ve. Don't change it now though!

If so then I would have 227.5+12-50= 189.5lbs for the vertical hip force. For the horizontal component, 394lbs=-Fhip. So Fhip=-394lbs.
Correct. Now combine them to give the magnitude and direction of Fhip.
 

Related Threads on How to solve for the moment of inertia and angular acceleration?

Replies
3
Views
2K
Replies
7
Views
5K
Replies
17
Views
5K
Replies
3
Views
4K
Replies
9
Views
4K
Replies
4
Views
19K
Replies
2
Views
2K
Replies
16
Views
11K
Replies
4
Views
2K
Replies
3
Views
2K
Top