Suggestions:
Draw an FBD of the ball. Find the moment about point O of the forces acting on the sphere. Note that this moment does not have a z-component. What does this mean? [The z-component of the angular momentum about point O is constant.] Use this to find the new angular velocity of the ball about the z-axis through O.
Let us know if you have questions on this.
14 comments:
I had a quick question. The r_dots are zero right? I came to this conclusion because an overhead view of the spinning system, the velocty (tangent to path) is only in the e_theta direction.
Also, i believe they are zero because you are looking at only final and initial states, not the in between. At the final and initial states, r is not changing.
I see that we need to take the moments about O to solve this problem. Does this mean that we neglect the normal and gravitational forces so that we are only left with tension and thus the sum total of the moments is zero?
Also, to find the work, I'm not getting the right answer. Shouldn't it just be T_2+V_2-T_1? (if I assume initially zero potential)
This then expands to:
(1/2)m(omega_2*r_2)^2+mgh_2-(1/2)m(omega_1*r_1)^2
Does this look right?
i am also having trouble finding the work. i cancelled out potential, but still cannot seem to come up with the book answer. can anyone steer me in the right direction?
I was having this problem also. But I realized, that since omega is about the z-axis, the r we're using to find v is not b, but rather b*cos(30). So for the kinetic energy terms, it's 1/2*m*(b*cos30*omega)^2.
I'm having trouble getting the correct answer for the first part of the question. I have H1=H2 so
.3cos(30)^2*(.2)*4=.2cos(30)^2*(.2)*w
when I solve for w I get 6rad/s. I'm not sure where i'm going wrong.
Your equation should actually be (.3cos(30))^2*(.2)*4=(.2cos(30))^2*(.2)*w
This way I am getting w = 9.
Is angular momentum consevred? I also have not been able to find the correct (omega) value.
John,
With the equation being H_o = r x (mv), if you write v in terms of polar coordinates, the cross product produces a r^2.
__ Haha, Sorry Sam, I deleted my post before I saw you had replied. I had figured it out by the time I refreshed the page. Thanks for your reply.
I have solved for w, so thank you for all of your suggestions. However, I am now solving for work and do not know what to use as my 'h' value in the potential equation. I know that you can use b*sin(30) to get the vertical component of 'b' but if the initial potential is zero then shouldn't 'h' be the distance from b*sin(30) to the ground? Please let me know any thoughts.
j.t.
I think V2-V1=mg(0.3sin30-0.2sin30)
with 0.2 being the new value of b.
Then you can use T2+V2-T1-V1=U
btw T1 is not zero. T1=0.5m(0.3cos30*4)^2
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