What Is the Minimum Speed of a Dart to Complete a Loop with a Lead Sphere?

  • Thread starter Thread starter yandao87
  • Start date Start date
  • Tags Tags
    Lead Sphere
AI Thread Summary
The discussion centers on calculating the minimum speed of a dart needed to ensure a lead sphere completes a circular loop after a collision. The lead sphere, hanging from a wire, is struck by a dart, and the key to solving the problem lies in understanding the conservation of momentum and energy. Initially, the user incorrectly calculated the dart's speed using the velocity at the top of the swing instead of the bottom, leading to an erroneous result. After clarifying the concepts of kinetic and potential energy, the user correctly determined the speed at the bottom of the swing and subsequently calculated the required initial speed of the dart. The final conclusion confirms that the initial speed of the dart must be approximately 48.95 m/s to achieve the desired outcome.
yandao87
Messages
10
Reaction score
0

Homework Statement



A 16.00 lead sphere is hanging from a hook by a thin wire 3.20 long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 5.50 steel dart that embeds itself in the lead sphere.

What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?


Homework Equations





The Attempt at a Solution



1) I've try equating T = mv^2 / r = mg when the sphere reaches the top of the swing.

v=5.6 as r = 3.2 and m cancels out.

2) then initial p = final p
5.5 v = (16+5.5) (5.6)
i'll get initial v of 21.9m/s for the dart but its wrong.

can anyone tell me what's wrong? or should i just the mgh PE at the top approach?
 
Physics news on Phys.org
One issue is that just getting to the top will not make the sphere travel in a complete circle. The lead sphere must reach the top with enough velocity so that its centrifugal force overcomes gravity. Otherwise the thin wire will buckle and the lead sphere drop before you get to the top.

Even though it has enough kinetic energy to reach the top not all of that will convert to potential energy and the path will look like the letter 'G' on its back .

There is a difference between the mass on a wire and a mass on a light rigid rod.
 
jambaugh said:
One issue is that just getting to the top will not make the sphere travel in a complete circle. The lead sphere must reach the top with enough velocity so that its centrifugal force overcomes gravity. Otherwise the thin wire will buckle and the lead sphere drop before you get to the top.

Even though it has enough kinetic energy to reach the top not all of that will convert to potential energy and the path will look like the letter 'G' on its back .

There is a difference between the mass on a wire and a mass on a light rigid rod.

ok, but any idea how i can approach this problem??
 
yandao87 said:
1) I've try equating T = mv^2 / r = mg when the sphere reaches the top of the swing.

v=5.6 as r = 3.2 and m cancels out.
Good! Be careful how you write it though. While mv^2/r = mg, that's not the tension. (I assume T stands for tension; T = 0.)

That's the minimum speed of the "sphere+dart" at the top of the swing. Now figure out the speed at the bottom of the swing, where the collision takes place. (What's conserved as the sphere swings up?)

2) then initial p = final p
5.5 v = (16+5.5) (5.6)
i'll get initial v of 21.9m/s for the dart but its wrong.
It's wrong because you're using the speed at the top instead of at the bottom.

jambaugh said:
One issue is that just getting to the top will not make the sphere travel in a complete circle. The lead sphere must reach the top with enough velocity so that its centrifugal force overcomes gravity. Otherwise the thin wire will buckle and the lead sphere drop before you get to the top.
This part was done correctly in the first step. (Except for the "T =" part.)
 
Doc Al said:
Good! Be careful how you write it though. While mv^2/r = mg, that's not the tension. (I assume T stands for tension; T = 0.)

That's the minimum speed of the "sphere+dart" at the top of the swing. Now figure out the speed at the bottom of the swing, where the collision takes place. (What's conserved as the sphere swings up?)


It's wrong because you're using the speed at the top instead of at the bottom.


This part was done correctly in the first step. (Except for the "T =" part.)

hi thanks thanks so much for the reply. i understand that for the first part, it should be T+mg = mv^2/r , but T= 0 as assumed.

but for the 2nd part,
2) then initial p = final p
5.5 v = (16+5.5) (5.6)
i'll get initial v of 21.9m/s for the dart but its wrong.

It's wrong because you're using the speed at the top instead of at the bottom.

i'm using v=5.6 at the top for final p. i don't get what you mean by that i should use the speed at the bottom instead. isn't that what I'm supposed to find(speed at the bottom)?
 
yandao87 said:
i'm using v=5.6 at the top for final p. i don't get what you mean by that i should use the speed at the bottom instead. isn't that what I'm supposed to find(speed at the bottom)?
You are supposed to find the speed of the dart before the collision. To do that, you need the speed of the "sphere + dart" after the collision. You can find that post-collision speed from the speed at the top of the swing.

What happens to the speed of the object as it swings up?
 
Doc Al said:
You are supposed to find the speed of the dart before the collision. To do that, you need the speed of the "sphere + dart" after the collision. You can find that post-collision speed from the speed at the top of the swing.

What happens to the speed of the object as it swings up?

oh! i kind of get what you meant. so now, i have post collision at top = 5.6m/s

so,
1) at the top, it has k.e. of 0.5(16+5.5)(5.6)^2 and p.e. of (16+5.5)(9.8)(3.2*2)
2) solve for v using k.e. at bottom of 0.5(16+5.5)v^2, p.e. =0
3) i'll get v at bottom of 12.52m/s
then
1) using p(initial) = p(final)
5.5 * v = 12.52(16+5.5)
v = 48.95 for initial speed of dart

hope I'm finally correct?? :)
:p you're so kind to offer your help. really appreciate your precise and prompt replies.
 
Perfectly correct! :approve:

(And you are most welcome.)
 
Doc Al said:
Perfectly correct! :approve:

(And you are most welcome.)

thank you so much once again. :)
 
  • #10
Why the initial momentum of the lead sphere= 0?
 
  • #11
glen_ky said:
Why the initial momentum of the lead sphere= 0?
Presumably it's just hanging there, initially at rest.
 
Back
Top