An object is formed by attaching a uniform, thin rod with a mass of mr = 6.54 kg and length L = 5.12 m to a uniform sphere with mass ms = 32.7 kg and radius R = 1.28 m. Note ms = 5mr and L = 4R.
1)What is the moment of inertia of the object about an axis at the left end of the rod?
1417.969664
2)If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 467 N is exerted perpendicular to the rod at the center of the rod?
.8431209992
3)What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)
4)If the object is fixed at the center of mass, what is the angular acceleration if a force F = 467 N is exerted parallel to the rod at the end of rod?
0
5)What is the moment of inertia of the object about an axis at the right edge of the sphere?
Ive been having troubles with this problems for several hours i also talk to a bunch of TAs and even they couldn't help me out. If any one could help me out I would appreciate it. Thanks.
1)
Each shape as a different moment of inertia formula, for the left end of the rod:
I = (1/3)*M*L^2
where m = mass of rod
I = (1/3)*6.54*5.12^2
I = 57.14
Now you have to consider the moment of the left end of the sphere too:
I = (2/5)*M*r^2
where m = mass of sphere
I = (2/5)*32.75*1.28^2
I = 21.46
but you also have to consider the distance from the center of mass too so:
I= 21.46 + 32.7*(6.54+1.28)^2
I = 2021.14
Answer:
M.I of rod + M.I of sphere
21.46+2021.14 = ??
2)
(L/2)*F = M <-- torque
L = length of rod
F = Force
Once you find M
angular acceleration= M/(21.46+2021.14 )
(its divided by what you get in the first problem)
4) #4 is correct
try #3 and #4 by yourself, if you have problems post here again.
Here are links to formulas for M.I of different shapes.. Hopefully your prof provides formula sheets for this..
http://www.engineersedge.com/mechanics_machines/mass_moment_of_inertia_equations_13091.htm
http://hyperphysics.phyastr.gsu.edu/hbase/mi2.html
http://hyperphysics.phyastr.gsu.edu/hbase/isph.html
Every number I have posted is correct. I'm just having problems with 3 and 5. I have access to all of those equations too but i still can't seem to find the right answer. I've been trying for hours and like I said no one i have talked to knows how to do it even my TAs.
I can definitely help you out with these problems. Let's go through each question step by step.
1) To find the moment of inertia of the object about an axis at the left end of the rod, we need to consider the individual moments of inertia of the rod and the sphere, and then add them together. The moment of inertia of a thin rod rotating about one end is given by the formula: I_rod = (1/3) * mass_rod * length_rod^2.
Given that the mass of the rod is mr = 6.54 kg and length of the rod is L = 5.12 m, we can calculate the moment of inertia of the rod: I_rod = (1/3) * 6.54 kg * (5.12 m)^2.
Next, we need to find the moment of inertia of the sphere, which is given by the formula: I_sphere = (2/5) * mass_sphere * radius_sphere^2.
Given that the mass of the sphere is ms = 32.7 kg and the radius of the sphere is R = 1.28 m, we can calculate the moment of inertia of the sphere: I_sphere = (2/5) * 32.7 kg * (1.28 m)^2.
Finally, we can add the two moment of inertias together to get the total moment of inertia of the object about the left end of the rod.
2) To find the angular acceleration when a force F = 467 N is exerted perpendicular to the rod at the center of the rod, we can use the formula: angular acceleration = net torque / moment of inertia.
The net torque is equal to the force multiplied by the perpendicular distance to the axis of rotation. In this case, the perpendicular distance is equal to half of the length of the rod, so that would be L/2 = 5.12 m / 2.
Plug in the values for force (F = 467 N) and moment of inertia (which we calculated in question 1), and you can calculate the angular acceleration using the given formula.
3) To find the moment of inertia of the object about an axis at the center of mass of the object, we can use the parallel axis theorem. This theorem states that if we know the moment of inertia about an axis passing through the center of mass and parallel to a certain axis, we can find the moment of inertia about any axis parallel to it by adding the mass of the object multiplied by the square of the distance between the two axes.
Since we're given that the center of mass is located halfway between the center of the sphere and the left edge of the sphere (which would be at a distance of R/2 from the center of the sphere), we can calculate the moment of inertia about an axis at the center of mass using the parallel axis theorem.
4) If the object is fixed at the center of mass, the line of action of the force will pass through the axis of rotation, so the torque applied by the force is equal to zero. According to Newton's second law for rotation (τ = I * α), the torque is equal to the moment of inertia multiplied by the angular acceleration. Since the torque is zero and we have the value for the moment of inertia (which we can calculate using the parallel axis theorem as mentioned in question 3), we can solve for the angular acceleration, which will be zero in this case.
5) To find the moment of inertia of the object about an axis at the right edge of the sphere, we can again use the parallel axis theorem. Since we already know the moment of inertia about the center of mass (which we calculated in question 3), we can add the mass of the object multiplied by the square of the distance between the center of mass and the right edge of the sphere to find the moment of inertia about that axis.
I hope this explanation helps you solve these problems. If you need any further clarification, please let me know!