Calculate the tension in a rope attached to a ball in circular motion

[MattDutra123]

Homework Statement

A solid iron ball of mass 770kg is used on a building site. The ball is suspended by a rope from a crane. The distance from the point of suspension to the centre of mass of the ball is 12m. The ball is pulled back from the vertical and then released. It falls through a vertical height of 1.6m and strikes a wall. Calculate the tension in the rope just before impact.

Relevant Equations

F (centripetal force) = mv^2/r

The solution to the problem simply states: "Use of mv^2/r = 2000. T = (2000 + 7500) = 9500N". I don't understand this solution. Nothing more is provided. I don't know how you are supposed to find the radius (in order to use the centripetal force formula) merely from the information provided. Also, why is the tension the sum of the centripetal force and weight (7500)?
Can someone please explain this to me?
The problem provides a sketch. I have included it in the attachments.

 

[Chestermiller]

Using conservation of energy, what is the tangential velocity of the ball when it strikes the wall? Have you drawn a free body diagram of the ball just before it strikes the wall, showing the forces acting on the ball?

 

[haruspex]

how you are supposed to find the radius

"The distance from the point of suspension to the centre of mass of the ball is 12m."
Will that distance change as the ball swings?

 

[MattDutra123]

Using conservation of energy, what is the tangential velocity of the ball when it strikes the wall? Have you drawn a free body diagram of the ball just before it strikes the wall, showing the forces acting on the ball?

I calculated it to be 5.6 m/s. Is this correct? Still, I don't know how to calculate the radius.
Apologies for the delay in answering.

 

[MattDutra123]

"The distance from the point of suspension to the centre of mass of the ball is 12m."
Will that distance change as the ball swings?

I don't know. The problem given doesn't mention it. Could we use this vertical distance as the radius (turning the triangle sideways)? Applying the formula with radius as 12 gives me the required answer of 2000 (2012.2666). Is this a correct way to approach the problem or is it a coincidence?
Apologies for the delay in answering.

 

[haruspex]

Could we use this vertical distance

It is not, initially, a vertical distance. The diagram shows it in the final position.
Do you understand that what you need is the radius of the arc through which the ball swings, not the radius of the ball? (You are expected to treat that as a point mass.)

 

[MattDutra123]

It is not, initially, a vertical distance. The diagram shows it in the final position.
Do you understand that what you need is the radius of the arc through which the ball swings, not the radius of the ball? (You are expected to treat that as a point mass.)

Yes, I understand this, but from the diagram what I see is that the vertical length of the rope holding the ball is 12m. It swings horizontally to hit the wall through an unknown distance. That is my interpretation of the diagram. I cannot see how the ball is swinging through a 12 meter arc.

 

[PeroK]

Yes, I understand this, but from the diagram what I see is that the vertical length of the rope holding the ball is 12m. It swings horizontally (?) to hit the wall through an unknown distance. That is my interpretation of the diagram. I cannot see how the ball is swinging through a 12 meter arc.

The ball is not moving horizontally. The ball is moving in a circular arc of radius 12m.

Perhaps your first problem is that the question setter expects you to see this as obvious. But, if you don't see it, then it's hard to explain.

 

[Chestermiller]

In the final vertical position of the ball, the radius is 12 m, which is the same as the length of the rope. If the rope is inextensible, its length in the initial (non-vertical) position of the ball is also 12 m. If the length of the rope doesn't change, what is the radius in the initial position of the ball?