Solving for Force Given Rotational Inertia and Mass

  • Thread starter MathewsMD
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In summary: Yes, I meant ƩFy = FN - Mg - m(g-a), but in the next line of the OP you wrote M (g + a), which is not so easily explained?
  • #1
MathewsMD
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A 0.70-kg disk with a rotational inertia given by MR2/2 (M) is free to rotate on a fixed horizontal axis suspended from the ceiling. A string is wrapped around the disk and a 2.0-kg mass (m) hangs from the free end. If the string does not slip, then as the mass falls and the cylinder rotates, the suspension holding the cylinder pulls up on the cylinder with a force of:

A. 6.9N
B. 9.8N
C. 16N
D. 26N
E. 29N
ans: B

My solution:

ma = mg - F
Fr = Ia/r
mg - ma = Ia/r2

mg - ma = Ma/2

a = (mg)/(m + 0.5M)

ƩFy = FN - Mg - Ma

FN = M (g + a) = ~9.77N

Please point out any errors since I really want to ensure I understood every process throughly and correctly. If you have anything to add, in terms of helpful steps or things to consider in general, please reply! :)
 
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  • #2
You were doing well until here:
MathewsMD said:
ƩFy = FN - Mg - Ma
Look at the free body diagram for the disk (cylinder?).
Where is there an Ma in there?
 
  • #3
[ignore/]
 
Last edited:
  • #4
zahbaz said:
"ƩFy = FN - Mg - Ma"

Another way to ask what haruspex is getting at:
What does ƩFy equal? Meaning, do we have acceleration or are the sum of y forces zero?

Sorry, I meant to add ƩFy = 0, isn't that a correct assumption?
 
  • #5
MathewsMD said:
Sorry, I meant to add ƩFy = 0, isn't that a correct assumption?

Yes, but you don't have the right contributors to ƩFy.
 
  • #6
Ohh... yeah, that's a totally correct assumption. I misread the scenario!

I got the same answer, 9.77... but not sure where you have the Ma term as haruspex mentioned. If you consider the mass is falling with F = ma, how much tension remains on the line? This is a downwards force on the cylinder.

ƩFy = FN - Mg - tension
 
  • #7
Okay, thank your for clearing that up. Yes, I used the wrong variable name but calculated it as ma was tension.
 
  • #8
MathewsMD said:
Okay, thank your for clearing that up. Yes, I used the wrong variable name but calculated it as ma was tension.
OK, so you meant ƩFy = FN - Mg - m(g-a), right?
But in the next line of the OP you wrote M (g + a), which is not so easily explained?
 

FAQ: Solving for Force Given Rotational Inertia and Mass

1. What is rotational inertia?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution of the object and its axis of rotation.

2. How do you calculate rotational inertia?

Rotational inertia is calculated by multiplying the mass of an object by the square of its distance from the axis of rotation. This is known as the parallel axis theorem and can be written as I = mr^2, where I is the rotational inertia, m is the mass, and r is the distance from the axis of rotation.

3. What is the relationship between rotational inertia and force?

The relationship between rotational inertia and force is described by the equation F = ma = Iα, where F is the net force, m is the mass, a is the linear acceleration, I is the rotational inertia, and α is the angular acceleration. This equation states that the greater the rotational inertia of an object, the more force is required to produce a given angular acceleration.

4. Can rotational inertia be changed?

Yes, rotational inertia can be changed by altering the mass distribution of an object or by changing its axis of rotation. For example, holding a weight farther away from your body will increase the rotational inertia compared to holding it closer to your body.

5. How is rotational inertia used in real-world applications?

Rotational inertia is used in many real-world applications, such as in the design of vehicles, machines, and sports equipment. It is also important in understanding the motion of celestial bodies and in the development of robotics and other technologies that involve rotating parts.

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