Solve Moment of Inertia Problem Involving Forces & Torque

In summary: So in summary, the conversation involved solving a problem involving rotational motion, specifically calculating the moment of inertia, total torque, angular acceleration, and initial acceleration of a rotating plate with applied forces and a fixed axis. The conversation also included some calculations using the parallel-axis theorem and Newton's second law for rotational motion.
  • #1
ubiquinone
43
0
Hi I've been doing more problems involving rotational motion. This question involves moments of inertia and torque which I'm not sure on how to solve. If anyone could help, that would be great. Thanks!

Question: The uniform 4.0 kg thin plate is made to rotate about a vertical axis that is perpendicular to the page and through the point indicated on the lower right corner. Forces are applied to the plate as indicated.
Code:
Diagram:
       F_2 
      /       F_1
     /        \
    /__________\
   |           |
   |           |
   |           |
 b |     x     |
   |           |
   |           |
   |___________|
   /     c     Axis
  /F_3
x - marks the center of mass
F_1 = 20.0 N at 30 degrees
F_2 = 25.0 N at 40 degrees
F_3 = 8.0 N at 70 degrees
c = 55.0 cm
b = 25.0 cm
Calculate a) the moment of inertia about the plate about the indicated axis, b) the total torque about the axis through the corner, c) the angular acceleration of the plate and d) the magnitude of the initial acceleration of the center of mass on the plate when the mass starts from rest.

Part a) I think you use the parallel-axis theorem:
[tex]\displaystyle I = I_{cm}+Md^2[/tex]
where [tex]\displaystyle I_{cm}=\frac{1}{12}M(c^2+b^2}=\frac{1}{12}(4.0kg)(.55^2+.25^2)m^2=0.122 kg\cdot m^2[/tex]
[tex]\displaystyle d^2=\left(\frac{c}{2}\right)^2+\left(\frac{b}{2}\right)^2=\left(\frac{0.55}{2}\right)^2+\left(\frac{0.25}{2}\right)^2=0.0912m[/tex]
Therefore, [tex]\displaystyle I=(0.122kg\cdot m^2) + (4.0kg)(0.0912m^2)=0.487kg\cdot m^2[/tex]

Part b) Do you just find the individual moments and sum them up?
If so, [tex]\displaystyle \tau_1 = r_1\times F_1=+(0.25)(20)\sin 30^o = 2.5 N\cdot m[/tex], [tex]\displaystyle \tau_2 = r_2\times F_2 = -(\sqrt{0.55^2+0.25^2})(25)\sin(180^0-\tan^{-1}(\frac{0.25}{0.55}))= -6.25 N\dot m[/tex]
and [tex]\displaystyle \tau_3 =r_3\times F_3 = +(0.55m)(8N)\sin 20^o =1.50 N\cdot m[/tex]
Total torque [tex]= (2.5-6.25+1.50)N\cdot m = -2.25 N\cdot m[/tex]

Part c) The angular acceleration of the plate, well can't we just use Newton's second law for rotational motion, [tex]\displaystyle \tau_{total}=I\alpha[/tex]
Then, [tex]\displaystyle \alpha=\frac{\tau_{total}}{I}=\frac{-2.25 N\cdot m}{0.487kg\cdot m^2}=-4.62 rad/s^2[/tex]

Part d) The magnitude of the initial acceleration of the center of mass on the plate when mass starts from rest. This could be found just by [tex]\displaystyle a=r\alpha[/tex] where [tex]r=d[/tex]?

If so, [tex]\displaystyle a=(-4.62rad/s^2)\sqrt{\left(\frac{0.55}{2}\right)^2+\left(\frac{0.25}{2}\right)^2}=-1.395m/s^2[/tex]
 
Last edited:
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  • #2
Your work looks right, I'm only not sure about (d), though. Someone else's oppinion would come in handy.
 
  • #3
Yes (d) looks right to me. Point X is rotating about the fixed axis with a known angular acceleration, so its linear accel. = radius times angular accel.
 

FAQ: Solve Moment of Inertia Problem Involving Forces & Torque

1. What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in rotational motion. It is similar to mass in linear motion and depends on an object's mass distribution and shape.

2. How is moment of inertia calculated?

Moment of inertia can be calculated by summing the products of each particle's mass and its squared distance from the axis of rotation. It can also be calculated using the parallel axis theorem or by using the inertia tensor for more complex shapes.

3. How are forces and torque involved in solving moment of inertia problems?

Forces and torque are involved in solving moment of inertia problems because they are the external influences that cause changes in an object's rotational motion. By calculating the forces and torque acting on an object, we can determine the moment of inertia.

4. What is the relationship between moment of inertia and angular acceleration?

Similar to how mass affects linear acceleration, moment of inertia affects angular acceleration. The larger the moment of inertia, the larger the torque needed to produce a given angular acceleration.

5. How can solving moment of inertia problems be useful in real-world applications?

Solving moment of inertia problems can be useful in real-world applications such as designing machines, vehicles, and structures that involve rotational motion. It can also be used in analyzing the stability and control of objects in motion, such as satellites and aircraft.

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