Circular Motion Problem: Finding Speed of a Swinging Ball at a Given Angle

[bcd201115]

Homework Statement

a .30kg ball attached to the end of a string swings in a vertical circle having a radius of 1.4m. At the instant when the string makes an angle of 30° above the horizontal, the tension in the string is 3N. What is the speed of the ball at this instant?

Homework Equations

Can anyone help me with this one?

The Attempt at a Solution

T-mgcos30°=m(v²/r)
then solve for velocity? Am i doing this right?

 

[collinsmark]

Homework Statement

a .30kg ball attached to the end of a string swings in a vertical circle having a radius of 1.4m. At the instant when the string makes an angle of 30° above the horizontal, the tension in the string is 3N. What is the speed of the ball at this instant?

Homework Equations

Can anyone help me with this one?

The Attempt at a Solution

T-mgcos30°=m(v²/r)
then solve for velocity? Am i doing this right?

Red mine.

Almost right. Draw a free body diagram to ensure your force components are being added correctly.

[Edit: I highlighted something else in red. See PeterO's response below. He gives some good advice here. But it looks to me like you were originally on the right track by summing all of the force components in the radial direction (parallel to the string). You just need to make sure the components are correct and that you are adding up in the right direction. (Again, a FBD might help this process.)]

 

[PeterO]

Homework Statement

a .30kg ball attached to the end of a string swings in a vertical circle having a radius of 1.4m. At the instant when the string makes an angle of 30° above the horizontal, the tension in the string is 3N. What is the speed of the ball at this instant?

Homework Equations

Can anyone help me with this one?

The Attempt at a Solution

T-mgcos30°=m(v²/r)
then solve for velocity? Am i doing this right?

I don't think it is as simple as that.

I would be adding the weight force (down) to the Tension (along the string - you can only pull with a string)

This will give a net force angled down from the string.

Resolve that force parallel and perpendicular to the string.

The component parallel to the string represents the centripetal force.

The component perpendicular to the string represents the force slowing the ball - remember the ball will slow as it rises through the circle, and speed up as descends again.

Note: the string makes an angle of 30^o above the horizontal, so the tension in the string is at 30^o below the horizontal. Once the force of gravity is added to that, the net force will be at, perhaps 60^o below the horizontal.