Two questions:
1. Should we use polar cord. (like in class)?
2. What is the best way to utilize the contact point of the wheel to solve the problem?
So far I used Sum of the forces = ma (I have a in polar cord. - Rdot and Rdoubledot and thetadot are zero) I have alpha in terms of N, F r, and m. I heard people in the help room suggest there are 5 unknowns and 5 equations: where did the 5th unknown come from?
9 comments:
The way it was explained to me, you use moments of the contact point, but you also use the moment of the center of gravity. Adding in F=ma you should get what you need. This problem sucks to do but teaches good fundamentals, I think...
Hope this helps.
Does anyone know what the moment of inertia of the quarter-circular plate is?
-John
I was wondering about the mass moment of inertia also. How would you go about finding it?
Here is a sketch of the equations that you will need:
Newton/Euler:
sum F_x = m*a_Gx
sum F_y = m*a_Gy
sum M_G = I_G*alpha
Kinematics (vector equation):
a_G = a_C + alpha x r_G/C
[With a_C = 0 (since omega = 0), the kinematics equation gives two scalar equations in terms of a_Gx, a_Gy and alpha.]
The above, therefore, is a set of five equations in terms of five unknowns: F, N, a_Gx, a_Gy and alpha.
The mass moment of inertia (MMI) of a FULL circular disk about its center O is (1/2)*m*r^2. Therefore, the MMI of the QUARTER circle about O is: I_O = (1/8)*m*r^2. To find the MMI of the quarter circle about its center of mass G is found from the parallel axis theorem: I_G = I_O - m*d^2, where d is the distance between O and G.
The choice of coordinates is up to you. However, I would suggest using Cartesian since that is the easiest to use for the kinematics equation above.
Hope that this helps.
Just to clarify: the accel. of the contact point C is zero because the velocity of the contact point is zero??
I was wondering about that as well. Maybe it should be a_C = alpha * r ?
Also, why is
I_G = I_O - m*d^2
and not
I_G = I_O + m*d^2 ?
-John
Always be careful NOT to assume that since V_c = 0, A_c = 0. This cannot be true for a rolling object.. or the object would not be able to roll at all.
That's what we're wondering about.
Krousgrill said:
"With a_C = 0 (since omega = 0)"
which implies that a_C is zero because omega is zero.
I don't see how this can be true. I think that a_C = alpha * r but I can't get the answer, just close to the answer.
-John
All good questions.
The thing to remember about a no-slip point such as C is that its velocity is zero and that the component of acceleration along the contact surface is zero (here, a_Cx = 0). If you write the vector acceleration for the wheel, you get:
a_C = a_O + alpha x r_C/O - omega^2*r_C/O
where O is the geometric center of the wheel. Balancing the y-components of this equation and recognizing that O moves only in the horizontal direction (a_Oy = 0) gives:
a_C = a_Cy = - omega^2*r
Since omega = 0, then a_C = 0.
Sorry that I originally wrote that statement without much proof. Professor Nauman had mentioned a couple times in lecture that the acceleration of a no-slip contact point is omega^2*r. I just assumed that you guys that talked about that in lecture.
Concerning the question on mass moment of inertia, recall that the parallel axis theorem says that:
I_A = I_G + m*d^2
where G is the cm and A is an arbitrary point. Therefore, if you know I_A and need to know I_G, you re-arrange the equation to get:
I_G = I_A - m*d^2
Hence the "-" sign.
[An easy way to keep this straight is that the mass moment of inertia about the center of mass is always SMALLER than the mass moment of inertia about any other point on the body.]
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