Video: Understanding the Forces at Work on an Object Undergoing Circular Motion

A ball of mass 𝑀 attached to a massless string travels at a constant speed along a vertical circular path, as shown in the figure. At which point is the magnitude of the tension greatest?

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Video Transcript

A ball of mass 𝑀 attached to a massless string travels at a constant speed along a vertical circular path, as shown in the figure. At which point is the magnitude of the tension greatest? Choice a) position E; choice b) position C; choice c) position A; choice d) position B; or choice e) position D.

In the problem, we’re told that the ball has a mass, 𝑀, that it is travelling at a constant speed and in a vertical circular path. We’re solving for the point at which the magnitude of the tension is the greatest. Because our ball is moving in a circle, we can start by analyzing the centripetal force.

The centripetal force is the net force on an object that is directed towards the center of the circular path. The equation for centripetal force can be derived from Newton’s second law. The centripetal force, 𝐹 subscript c, is equal to the mass, π‘š, times the centripetal acceleration, π‘Ž subscript c. The equation for centripetal acceleration is π‘Ž subscript c equals 𝑣 squared divided by π‘Ÿ, where 𝑣 is our tangential velocity and π‘Ÿ is the radius of our circular path. Combining these two equations together, we get an expression for centripetal force in terms of mass, velocity, and radius. The equation states that the centripetal force felt by the object is equal to the mass of the object times the tangential velocity squared divided by the radius of the circular path.

Going back to the important information that we highlighted previously in our problem, we can see that at each position, the mass of the object stays the same at 𝑀. We were told that the speed is constant. And since it’s travelling in a circular path, the radius is the same at each position as well. This means that at each position, A, B, C, D, and E, our centripetal force is a constant value. Going back to the definition of centripetal force, we must now analyze all of the forces acting on the ball at each position.

The first force that we can draw in for each position will be the force of gravity. At each position, we have drawn a yellow arrow that we have labelled 𝐹 subscript g to represent the force of gravity. It is pointing straight down towards the center of the earth. The other force that’s acting on the ball at each position is the tension. The tension is the force within the string that’s attached to the ball as it goes around the circle. Therefore, the tension will point towards the center of the circle at each position. We have drawn a pink arrow at each position and labelled it 𝑇 to represent the tension.

Looking at the diagram, we can see that at position A and E, both tension and force of gravity are parallel to each other along the line that goes through the center of the circle. These are our extreme cases for our circle, meaning that one will have the greatest tension and one will have the least tension. Because the force of gravity and tension forces at positions B, C, and D are not parallel to each other, none of these choices will have the greatest tension. Therefore, we can cancel them out as possible choices. This means that answer choice b, position C; answer choice d, position B; and answer choice e, position D are all eliminated.

Now, we must determine whether position A or position E has the greatest tension. To do this, we must write the equations at both position A and position E for the centripetal force as a net force of both tension and force of gravity. At position E, both tension and force of gravity are pointed towards the center of the circle. Therefore, we add 𝑇 plus 𝐹 g to get our centripetal force. To isolate the tension, we must subtract force of gravity from both sides of the equation, thereby cancelling it out on the right side of the equation. Therefore, the tension at position E is equal to the centripetal force minus the force of gravity or weight of the ball.

At position A, we can see from the diagram that the tension is pulling towards the center and the force of gravity is pulling in the opposite direction. Therefore, our net force equation becomes centripetal force equals tension minus force of gravity. To isolate the tension, we must add the force of gravity to both sides of the equation, thereby cancelling it out from the right side of the equation. This leaves us with the equation for tension at position A as the centripetal force plus the force of gravity or weight of the ball.

The weight of the object has not changed between position E and position A. And as shown earlier, the centripetal force is also the same at both positions. Therefore, since the tension at position A is equal to adding the centripetal force plus the force of gravity. And at position E, the tension is equal to subtracting the centripetal force and the force of gravity. Position A will have the greatest tension.

The point at which the magnitude of the tension is the greatest on our circular path is that answer choice c, position A.

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