- #1
Brianjw
- 40
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Well I'm back. I've done most of everything, except for the last part of one problem and one complete problem that has me stumped.
First the problem I almost have solved:
While visiting friends at Cal State Chico, you pay a visit to the Crazy Horse Saloon. This fine establishment features a 200 kg mechanical bucking bull, that has a mechanism that makes it move vertically in simple harmonic motion. Whether the "bull" has a rider or not, it moves with the same amplitude (0.250 m) and frequency (1.50 Hz). After watching other saloon patrons hold on to the "bull" while riding, you (mass 75.0 kg) decide to ride it the macho way by not holding on. No one is terribly surprised when you come out of the saddle.
What is your speed relative to the saddle at the instant you return?
Now, what I got was that you're .11m above the equilibrium point at the time you leave the seat with an upward velocity of 2.11 m/s. I used what is given to find the phase angle.
[tex] \phi = arctan(-2.11/(1.5* 2*\pi*.11)) [/tex] which gives me:
[tex] \phi = -1.114 radians [/tex] *edit* was in degrees
I got omega = 1.5 * 2* pi
I then use x = A*Cos(omega*t+ phi) I was given that the time in the air was .538 seconds.
I think I'm close, but I can't seem to get the right answer. For the person, I used [tex] V = V_0 + a*t [/tex] which gave me -3.1624 m/s But when I try to find the velocity of the bull by using v = -omega * A *sin(omega *t + phi).
Anyone see what I'm doing wrong? Am I even close?
The next problem is a bit simpler but it seems to be going over my head:
A holiday ornament in the shape of a hollow sphere with mass 1.10×10^(-2)and radius 4.50×10^(-2) is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum.
Calculate its period. (You can ignore friction at the pivot. The moment of inertia of the sphere about the pivot at the tree limb is 5/3 M*R^2.)
I'm trying this approach cause none of the other ones would seem to work, too many unknowns.
[tex] \omega^2 = m*g*d/I [/tex]
it seems simple since everything is given except d. I just can't seem to eliminate d or solve for it in any way. Any ideas?
Thanks,
Brian
First the problem I almost have solved:
While visiting friends at Cal State Chico, you pay a visit to the Crazy Horse Saloon. This fine establishment features a 200 kg mechanical bucking bull, that has a mechanism that makes it move vertically in simple harmonic motion. Whether the "bull" has a rider or not, it moves with the same amplitude (0.250 m) and frequency (1.50 Hz). After watching other saloon patrons hold on to the "bull" while riding, you (mass 75.0 kg) decide to ride it the macho way by not holding on. No one is terribly surprised when you come out of the saddle.
What is your speed relative to the saddle at the instant you return?
Now, what I got was that you're .11m above the equilibrium point at the time you leave the seat with an upward velocity of 2.11 m/s. I used what is given to find the phase angle.
[tex] \phi = arctan(-2.11/(1.5* 2*\pi*.11)) [/tex] which gives me:
[tex] \phi = -1.114 radians [/tex] *edit* was in degrees
I got omega = 1.5 * 2* pi
I then use x = A*Cos(omega*t+ phi) I was given that the time in the air was .538 seconds.
I think I'm close, but I can't seem to get the right answer. For the person, I used [tex] V = V_0 + a*t [/tex] which gave me -3.1624 m/s But when I try to find the velocity of the bull by using v = -omega * A *sin(omega *t + phi).
Anyone see what I'm doing wrong? Am I even close?
The next problem is a bit simpler but it seems to be going over my head:
A holiday ornament in the shape of a hollow sphere with mass 1.10×10^(-2)and radius 4.50×10^(-2) is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum.
Calculate its period. (You can ignore friction at the pivot. The moment of inertia of the sphere about the pivot at the tree limb is 5/3 M*R^2.)
I'm trying this approach cause none of the other ones would seem to work, too many unknowns.
[tex] \omega^2 = m*g*d/I [/tex]
it seems simple since everything is given except d. I just can't seem to eliminate d or solve for it in any way. Any ideas?
Thanks,
Brian
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