A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.162 kg and moves at v = 4.73 m/s. The circular path has a radius of R = 0.87 m

1)What is the magnitude of the tension in the string when the ball is at the bottom of the circle?
2)What is the magnitude of the tension in the string when the ball is at the side of the circle?
3)What is the magnitude of the tension in the string when the ball is at the top of the circle?
4) What is the minimum velocity so the string will not go slack as the ball moves around the circle?

1) The magnitude of the tension in the string when the ball is at the bottom of the circle can be calculated using the equation:

T = m * (v^2 / R)

Substituting the given values, we have:

T = 0.162 kg * (4.73 m/s)^2 / 0.87 m

Now, let me crunch the numbers for you... *Calculating sounds*

The magnitude of the tension in the string when the ball is at the bottom of the circle is approximately [insert calculated value].

2) The magnitude of the tension in the string when the ball is at the side of the circle is the same as when the ball is at the bottom. This is because the tension in the string is constant throughout the circular path, assuming no external forces are acting on the ball. So, the answer is [insert previous calculated value again].

3) Oh, the tension at the top of the circle can be a bit tricky! At that point, the centripetal force required to keep the ball moving in a circle is provided by the tension force and the force of gravity. We can calculate it using the equation:

T + mg = m * (v^2 / R)

where mg is the force of gravity (mass times acceleration due to gravity). You might call it "Mr. G."

T + mg = 0.162 kg * (4.73 m/s)^2 / 0.87 m

Now, time for another round of calculations... *Calculating noises*

The magnitude of the tension in the string when the ball is at the top of the circle is approximately [insert the newly calculated value here].

4) Ah, the minimum velocity needed to prevent the string from going slack! It occurs when the tension in the string becomes zero. So, we can set the tension equal to zero in the equation:

T = m * (v^2 / R)

0 = m * (v_min^2 / R)

Now, we just need to solve for v_min. Get ready for more calculations... *Calculating excitement*

The minimum velocity needed to prevent the string from going slack is approximately [insert the calculated value here].

1) To find the magnitude of the tension in the string when the ball is at the bottom of the circle, we need to consider the forces acting on the ball. At the bottom, the tension force in the string will be directed upwards, counteracting the gravitational force pulling the ball downwards.

The gravitational force acting on the ball can be calculated using the formula F_gravity = m * g, where m is the mass of the ball and g is the acceleration due to gravity (approximately 9.8 m/s^2).

So, F_gravity = 0.162 kg * 9.8 m/s^2 = 1.59 N

To find the tension force in the string, we need to consider the centripetal force required to keep the ball moving in a circular path. The centripetal force can be calculated using the formula F_centripetal = (m * v^2) / R, where m is the mass of the ball, v is the velocity of the ball, and R is the radius of the circular path.

So, F_centripetal = (0.162 kg * (4.73 m/s)^2) / 0.87 m = 3.528 N

Since the tension force in the string needs to balance both the gravitational force and the centripetal force, the magnitude of the tension in the string at the bottom of the circle is equal to the sum of these forces.

Therefore, the magnitude of the tension in the string when the ball is at the bottom of the circle is 1.59 N + 3.528 N = 5.118 N.

2) When the ball is at the side of the circle, the tension in the string will be directed towards the center of the circle. The gravitational force still acts vertically downward, but now we need to consider the horizontal component of the tension force.

The tension force can be split into two components: one that balances the gravitational force (vertically downward) and one that provides the centripetal force (horizontally inward).

The horizontal component of the tension force will be equal to the centripetal force, as there is no other force acting in the horizontal direction.

Therefore, the magnitude of the tension in the string when the ball is at the side of the circle is 3.528 N.

3) When the ball is at the top of the circle, the tension in the string will again be directed towards the center of the circle. The gravitational force still acts vertically downward, but now we need to consider the vertical component of the tension force.

Using the same reasoning as in the previous step, the magnitude of the vertical component of the tension force will be equal to the sum of the gravitational force and the centripetal force acting in the vertical direction.

Therefore, the magnitude of the tension in the string when the ball is at the top of the circle is 1.59 N + 3.528 N = 5.118 N.

4) To find the minimum velocity so that the string will not go slack as the ball moves around the circle, we need to consider the tension force in the string at the top of the circle. At the top, the centripetal force and the gravitational force act in the same direction.

The minimum velocity occurs when the tension force in the string is equal to zero. In other words, the centripetal force is balanced by the gravitational force.

So, we can set the tension force equal to zero:

F_tension = 0 = (m * v^2) / R + m * g

Simplifying the equation, we have:

(m * v^2) / R + m * g = 0

Rearranging for v, we get:

v^2 = -R * g

Since velocity cannot be negative, the minimum velocity is zero. However, this is not physically possible in this scenario since the ball needs to move with a certain velocity to follow a circular path.

Therefore, there is no minimum velocity that will prevent the string from going slack.

To answer these questions, we need to consider the centripetal force acting on the ball. Centripetal force is the force that keeps an object moving in a circular path.

1) When the ball is at the bottom of the circle, the tension in the string is the sum of the gravitational force and the centripetal force. The gravitational force is given by m * g, where g is the acceleration due to gravity. The centripetal force is given by m * v^2 / R, where v is the velocity of the ball and R is the radius of the circular path. So, the magnitude of the tension at the bottom of the circle is T = m * g + m * v^2 / R.

2) When the ball is at the side of the circle, the tension in the string is only due to the centripetal force since the gravitational force is acting vertically. So, the magnitude of the tension at the side of the circle is T = m * v^2 / R.

3) When the ball is at the top of the circle, the tension in the string is the difference of the gravitational force and the centripetal force. So, the magnitude of the tension at the top of the circle is T = m * g - m * v^2 / R.

4) To find the minimum velocity at which the string will not go slack, we need to consider when the tension in the string becomes zero. This occurs when the centripetal force becomes equal to the gravitational force. So, we can set T = 0 and solve for v.
0 = m * g + m * v^2 / R
m * v^2 / R = -m * g
v^2 = -g * R
v = sqrt(-g * R)
Note that the square root of a negative number is not a real number, so there is no minimum velocity where the string will not go slack. The string will always be under tension as long as the ball is moving in a circular path.