Calculate Velocity of Mechanical Bucking Bull at Cal State Chico

[marekkpie]

Homework Statement

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 1.61 m and frequency 0.474 Hz. Being from Texas you decide to ride it the “macho” way by NOT holding on. To no ones surprise you go flying out of the saddle. While waiting for your bruises and pride to heal, you decide to calculate how fast upward you were moving when you left the saddle.

Give your answer in m/s to the second decimal place.

Homework Equations

1. 1/2\,m{v}^{2}+1/2\,k{x}^{2}=1/2\,k{A}^{2}
2. v=\sqrt {{\frac {k}{m}}}\sqrt {{A}^{2}-{x}^{2}}
3. v=\omega\,A
4. \omega=2\,\pi\,f

The Attempt at a Solution

I had several attempts, all of which get me in the right ballpark, but not the correct answer, which is 3.49 m/s.

1) Using the third and fourth equations:
v=2\,\pi\,fA
= 4.79.
After thinking about it some more, this equation won't work because this finds the maximum velocity while still connected to the spring.

2) I attempted to use the second equation, but again, this would mean I was still attached to the bull, and thus x = A, making v = 0.

3) I reread the problem and saw that it said AFTER I left the bull, what was my speed. So I thought, conservation of energy:

1/2\,k{A}^{2}=1/2\,m{v}^{2}+mgh, with h = A.

then solving for v:

v=A\sqrt {{\frac {k}{m}}-2\,gA}, but I still needed to get rid of the k, so I thought:

\omega=\sqrt {{\frac {k}{m}}}, but this is where I am almost certain that my leap of faith isn't true; however, I continued, using the fourth equation to substitute for \omega, and got -.82 m/s...

So, I'm stuck. I think I'm on the right track with conservation of energy, but I don't think I can create \omega into that equation.

 

[Doc Al]

Hint: What's the value of the bull's acceleration when you begin to lose contact?

 

[marekkpie]

Hrmm...I can't quite think of how to use that at the moment, but I did remember about the equation for displacement in SHM:

x=A\cos \left( \omega\,t+\phi \right)

And the first derivative of this equation gets me:

-A\sin \left( \omega\,t+\phi \right) \omega

The problem is I can't seem to find the correct t for when the rider gets launched off the bull.

Since I don't have any "seconds" given, I have been using the fact that:

\omega\,T=2\,\pi

By just brute force, I have narrowed down that:

6\,t<T AND T<7\,t, but i have a feeling I am making it too complicated.

Again, maybe I've just fried my brain for too long today, but I can't find a great use of the bull's acceleration at the moment.

 

[Doc Al]

Hrmm...I can't quite think of how to use that at the moment, but I did remember about the equation for displacement in SHM:

x=A\cos \left( \omega\,t+\phi \right)

And the first derivative of this equation gets me:

-A\sin \left( \omega\,t+\phi \right) \omega

OK, that's the velocity, so you'll definitely need that. (You'll also need the second derivative, which will give you the acceleration.)

The problem is I can't seem to find the correct t for when the rider gets launched off the bull.

Since I don't have any "seconds" given, I have been using the fact that:

\omega\,T=2\,\pi

By just brute force, I have narrowed down that:

6\,t<T AND T<7\,t, but i have a feeling I am making it too complicated.

You need to specify the physical condition that holds when the rider is thrown off. That's where acceleration comes in. A few more hints: Since you are not attached to the bull, what's the maximum downward acceleration you can have? What happens to the normal force when you begin to lose contact?

Also: Don't bother finding the actual time at which you get tossed off; all you care about is the speed.