Solving Momentum Questions: Bird on a Swing

  • Thread starter Thread starter lvuittongirl22
  • Start date Start date
  • Tags Tags
    Momentum
AI Thread Summary
Momentum conservation is key to solving the problem of the bird on the swing. When the bird takes off horizontally at 2.00 m/s, the swing will move backward, and its tangential velocity can be used to determine how high it rises. The kinetic energy of the swing converts to potential energy at its highest point, allowing the calculation of height using the equation (1/2)mv^2 = mgh. The discussion highlights the challenge of understanding physics concepts without effective teaching support. Overall, the exchange emphasizes the importance of grasping fundamental principles to solve momentum-related questions.
lvuittongirl22
Messages
5
Reaction score
0
Hey Everyone! I don't quite understand how to figure this out...Any help would be greatly appreciated!
The bird perched on the swing has a mass of 52.0 g and the base of the swing has a mass of 153 g. The swing and the bird are originally at rest, and then the bird takes off horizontally at 2.00 m/s. How high does the base of the swing rise above the original level? Disregard friction.
Thanks! ~*~Jazzy~*~
 
Physics news on Phys.org
Well, momentum is conserved, so find the velocity the swing will move back with.

That would be the tangential velocity of the swing's circular motion because it is attached to the string.

Now, figure out how you can use that tangential velocity to find out how high the swing would go if you treat it like a pendulum.
 
how exactly do I use the tangential velocity to find the height?? Sorry if I sound like I'm dumb or something, but I don't really get physics too well...
 
I really don't remember, I expected you to look it up. :) Let's see . . .

How about this. You know the kinetic energy. You know the mass. You know gravity. The potential energy at it's top hight will be mgh. Since there is no friction, all the kinetic energy will turn into potential energy. Set (1/2)mv^2 equal to mgh and solve for h.

Now there's nothing really left for you to do. :( I'm sure your happy though. ;)
 
thanks - i just don't understand physics - if i would have known that i was going to have problems, i never would have taken it - my main problem is that my teacher doesn't teach - he just kinda presents the material, and i don't get it, so i have to try to teach myself, which is really hard when you have no inkling of what to do...but Thank you so much for helping direct me on the right path! You have no idea how much it means to me!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top