How to Calculate Tension in a Vertical Circle and Accelerating System

[LostShadow]

Hello. Got a few simple physics problems.

Homework Statement

Cannon fires a ball at 30 degrees to the horizontal at a speed of 50 m/s. The cannon is 10 m above the ground. What is the ball's speed when it eventually falls to the ground, using energy conservation! That really threw me off.

Homework Equations

t = square root of ( 2 times distance / 9.8 m/s^2 )
= square root of ( 2(10 m) / 9.8 m/s^2 )
= 1.42 s

d = 50 m/s (1.42) = 71 m in the horizontal

The Attempt at a Solution

So I know v = at, but I don't know the use "energy of conservation" thing.

Homework Statement

A .1 kg ball is at the end of 1.0 m string. It is swung in a vertical circle whose center is 2.0 m above floor. When the string is horizontal and the ball is moving upward, the string suddenly breaks, and the ball reaches a height of 5.0 m. What was the tension in the string the instant it broke?

Homework Equations

Tension = N - mg that I know of.

From what I know, tension is another cute word for force, right, like T = ma<something theta>, where something = {sine, cosine, etc. }

The Attempt at a Solution

Is this an L = mvr kind of problem? I have the mass of ball, velocity of ball, and radius of circle, but don't know how to convert it to tension.





And I have another question.

Suppose I have something like.

____|____
|________|
____|____
|________|


Where | is a string, and the 2 blocks are accelerating upward at 5 m/s^2.

How do I find the tension at the top of the rope? The blocks are 2.0 kg each, and the rope is .5 kg.

The force = sum of blocks and rope * acceleration.

If the whole thing is going upward, do I still take into consideration the 9.8 m/s^2?

 

[Shooting Star]

1. You know the ke + pe initially. Equate this to the final ke + pe. It's easy.

2. Given the initial vertical velo v upward, you can find the h-max and vice versa. So, you get v. Now to find T, equate centripetal force to T.

3. Motion is due to tension only. Use F=ma. If this is taking place on earth, then add F-mg =ma.

 

[PhanthomJay]

You should post each question separately, as it makes it easier to respond. For number one, since only gravity is acting, the conservation of energy principle states that the sum of the initial potential and kinetic energy of the system must be equal to the sum of the final potential and kinetic energy of the system.

 

[LostShadow]

1. You know the ke + pe initially. Equate this to the final ke + pe. It's easy.

Okay, so should the initial KE + PE be before the cannon fired, or at peak?

So I don't have the mass, can I take them out on both sides?

Initial = final.
1/2(v^2) + gh = 1/2(v^2) + gh.

Otherwise I get:

1/2(m)(80 m/s^2) + m(-9.8 m/s^2)10 m = PE + KE again.

2. Given the initial vertical velo v upward, you can find the h-max and vice versa. So, you get v. Now to find T, equate centripetal force to T.

Okay how do I find initial velocity?

3. Motion is due to tension only. Use F=ma. If this is taking place on earth, then add F-mg =ma.

Alright, thanks.

 

[Shooting Star]

Have you solved it or not? I couln't understand that from your thanks.

 

[LostShadow]

The alright, thanks was just for number 3. :/

 

[Shooting Star]

1. final ke + pe at ground = initial ke + pe at cannon. Take mass m. PE at ground is 0.

2. initial velo v= velo upward when string is horizontal. mv^2/r = T. h is given. From that find v. Then find T.

Clear?

 

[LostShadow]

1. final ke + pe at ground = initial ke + pe at cannon. Take mass m. PE at ground is 0.

So (1/2)m(50 m/s^2) + m(-9.8 m/s^2)(10 m) = (1/2)m(new velocity^2) + 0

= 25 m/s^2 * m + -98 m^2 * m/s^2 = (1/2)m(new velocity^2) + 0

Okay, so I guess I don't need the unknown m, can I get rid of it? By diving it by both sides? And just use:

1/2 v^2 + gh = 1/2 v^2 + 0?

Then I'll solve for final velocity.

2. initial velo v= velo upward when string is horizontal. mv^2/r = T. h is given. From that find v. Then find T.

Clear?

Er, so I find v before T. But I don't know T. T = mv^2/r. I have 2 unknowns, v and T.

Sorry, still a bit lost.

 

[Shooting Star]

So (1/2)m(50 m/s^2) + m(-9.8 m/s^2)(10 m) = (1/2)m(new velocity^2) + 0

= 25 m/s^2 * m + -98 m^2 * m/s^2 = (1/2)m(new velocity^2) + 0

Okay, so I guess I don't need the unknown m, can I get rid of it? By diving it by both sides? And just use:

1/2 v^2 + gh = 1/2 v^2 + 0?

Then I'll solve for final velocity.

Somewhat there. The two velocities are different.

The unit of speed is m/s, not m/s^2.

The energy eqn is: mv1^2/2 + mgh = mv2^2/2 + mg*0.

Er, so I find v before T. But I don't know T. T = mv^2/r. I have 2 unknowns, v and T.

(If you have already found the value of v, then how is it an unknown?)

You know the height to which it rises. From there you find v. Then plug in the value of v to find T.

If h is the height to which it rises from where it had vertical velo v, then v^2 = 2gh. Can you do the rest?

 

[LostShadow]

(If you have already found the value of v, then how is it an unknown?)

I don't have velocity.

?

 

[Shooting Star]

I don't have velocity.

?

You had it. Anyway, now I've given you the formula. Read my last post.