Question

A ball that has a mass of \(1.00 \mathrm{~kg}\) is attached to a string \(1.00 \mathrm{~m}\) long and is whirled in a vertical circle at a constant speed of \(10.0 \mathrm{~m} / \mathrm{s}\). a) Determine the tension in the string when the ball is at the top of the circle. b) Determine the tension in the string when the ball is at the bottom of the circle c) Consider the ball at some point other than the top or bottom. What can you say about the tension in the string at this point?

Short answer

Short Answer: The tension in the string when the ball is at the top of the circle is 109.81 N, and when the ball is at the bottom of the circle, it is 90.19 N. At other points in the vertical circle, the tension varies depending on the angle, being greater than just the gravitational force and providing the centripetal force needed for circular motion.

Step-by-step solution

1. Identify forces acting on the ball at the top of the circle.

At the top of the circle, there are two forces acting on the ball: gravitational force (downward) and tension force (upward).

2. Apply Newton's second law for the vertical forces at the top of the circle.

We can write the equation for the net force acting on the ball at the top of the circle: \(T - mg = m \dfrac{v^2}{r},\) where \(T\) is the tension in the string, \(m\) is the mass of the ball, \(g\) is the acceleration due to gravity, \(v\) is the speed of the ball, and \(r\) is the radius of the circle.

3. Solve the equation for tension at the top of the circle.

Substitute the given values into the equation: \(T - (1.00~\mathrm{kg})(9.81~\mathrm{m/s^2}) = (1.00~\mathrm{kg})\dfrac{(10.0~\mathrm{m/s})^2}{1.00~\mathrm{m}}.\) Now, solve for \(T\): \(T = 1.00~\mathrm{kg} \cdot 9.81~\mathrm{m/s^2} + 1.00~\mathrm{kg} \cdot 10.0~\mathrm{m^2/s^2} = 9.81~\mathrm{N} + 100.0~\mathrm{N} = 109.81~\mathrm{N}\). #a) Answer:# The tension in the string when the ball is at the top of the circle is 109.81 N. #b) Determine the tension in the string when the ball is at the bottom of the circle.#

1. Identify forces acting on the ball at the bottom of the circle.

At the bottom of the circle, there are two forces acting on the ball: gravitational force (downward) and tension force (upward).

2. Apply Newton's second law for the vertical forces at the bottom of the circle.

We can write the equation for the net force acting on the ball at the bottom of the circle: \(T + mg = m \dfrac{v^2}{r}.\)

3. Solve the equation for tension at the bottom of the circle.

Substitute the given values into the equation: \(T + (1.00~\mathrm{kg})(9.81~\mathrm{m/s^2}) = (1.00~\mathrm{kg})\dfrac{(10.0~\mathrm{m/s})^2}{1.00~\mathrm{m}}.\) Now, solve for \(T\): \(T = 1.00~\mathrm{kg} \cdot 10.0~\mathrm{m^2/s^2} - 1.00~\mathrm{kg} \cdot 9.81~\mathrm{m/s^2} = 100.0~\mathrm{N} - 9.81~\mathrm{N} = 90.19~\mathrm{N}\). #b) Answer:# The tension in the string when the ball is at the bottom of the circle is 90.19 N. #c) Consider the ball at some point other than the top or bottom. What can you say about the tension in the string at this point?# #c) Answer:# As the ball moves around the circle, the net force continuously changes direction and magnitude. The tension in the string will be greater than just the gravitational force, as it has to provide the centripetal force needed for the circular motion as well. Although the centripetal force remains constant due to the constant velocity, the tension in the string varies depending on the angle in the vertical circle.