What is the speed and angle of a swinging pendulum?

[Ready2GoXtr]

Homework Statement

1) A pendulum consists of a 2 kg stone swinging on a 4 m long string
of negligible mass. The stone has a speed of 8 m/s when it passes its
lowest point. a) What is the speed of the rock when it is at an angle of
60 degrees to the vertical? b) What is the greatest angle with the vertical
that the string will make during the stone's motion?

Homework Equations

Not really sure Just trying to do it with energy conservation..

The Attempt at a Solution

Well I figured out that at 60 degree's the height of the stone from the ground is 2m

Thne I figured I should use E = k+mgy ...
and came up with Einitial = (1/2)(2kg)(8m/s)^2 + (2kg)(-9.8m/s^2)(4m)

Efinal = (1/2)(2kg)(v)^2 + (2kg)(-9.8m/s^2)(2m)


Where to go Next... no clue

 

[Dick]

"energy conservation". Set Einitial=Efinal and solve for v.

 

[Dick]

Be careful about the sign and stuff on your gravitational terms though. It's mgh, where h is the height above the ground and there is no minus sign on the g.

 

[Ready2GoXtr]

"energy conservation". Set Einitial=Efinal and solve for v.

Really... can you tell me what that is derived from like I've seen deltaE = E final - E initial..

 

[Dick]

Really... can you tell me what that is derived from like I've seen deltaE = E final - E initial..

There's no energy going into or out of the system. So deltaE=0. Right?

 

[Ready2GoXtr]

Im getting v @ 60 degrees = sqrt(55.2) You?

 

[Ready2GoXtr]

Okay for part B I am going to assume that Einitial will remain the same... But Efinal will be = 0 + (2kg)(9.8m/s^2)(4m(1-cos(theta))

 

[Ready2GoXtr]

Then again I am assuming I can set them equal to each other.. cause no energy is gained or lost. so That gives me 94.4 = (2kg)(9.8m/s^2)(4m(1-cos(theta))

94.4 = 19.6(4m(1-cos(theta))

94.4 = 78.4(1-cos(theta))

.204081633 = -cos(theta)

-.204081633 = cos(theta)

cos^(-1)(-.2040816330) = theta

 

[Ready2GoXtr]

im getting (theta) = 101.775744 degrees

101.8 Degrees

 

[Dick]

Im getting v @ 60 degrees = sqrt(55.2) You?

Sorry. No, I don't get that. Can you post details? And you didn't post part B either. So I don't know what you are working on.