A .275 kg. object is swung in a vertical circular path on a string .850 m. long. A) What are the forces acting on the ball at any point along this path? B) Draw free-body diagrams for the ball when it is at the bottom of the circle and when it is at the top. C) If its speed is 5.2m/s at the top of the circle, what is the tension in the string there? D) If the string breaks when its tension exceeds 22.5 N. what is the maximum speed the object can have at the bottom before the string breaks?

d) F = ma (in the 'y' direction)

*ball at bottom mean 'T' is upward and 'mg' is downward*

T - mg = ma
T - mg = m(v^2/r)
(r(T - mg))/m = v^2
v = sqrt((r(T - mg))/m)

v = sqrt((.85(22.5 - (.275*9.8))/.275)

v = 7.82 m/s

m g down

Tension of string toward center
At top
m g + T = m v^2/r
T = m (v^2/r-9.81
T = .275(5.2^2/.85 -9.81)
= 6.05 N

At bottom
T - m g = m v^2/r
22.5 = .275 (v^2/.85 + 9.81)

need a fee body diagram

A) The forces acting on the ball at any point along this path are gravitational force (weight), tension force, and centripetal force.

B) At the bottom of the circle, the free-body diagram would include the following forces:
- Gravitational force (downward)
- Tension force (upward)
- Centripetal force (towards the center of the circle)

At the top of the circle, the free-body diagram would include the following forces:
- Gravitational force (downward)
- Tension force (downward)
- Centripetal force (towards the center of the circle)

C) To determine the tension in the string at the top of the circle, we need to find the net force acting on the object. The net force is given by:

Net force = mass * acceleration

Since the object is moving in a circle, the acceleration is the centripetal acceleration:

Centripetal acceleration = (velocity squared) / radius

Given:
- Mass (m) = 0.275 kg
- Velocity (v) = 5.2 m/s
- Radius (r) = 0.850 m

Plugging in these values, we can calculate the centripetal acceleration:

Centripetal acceleration = (5.2^2) / 0.850 = 31.948 m/s^2

Now, let's find the net force:

Net force = mass * acceleration = 0.275 kg * 31.948 m/s^2 = 8.785 N

Since tension is the centripetal force, the tension in the string at the top is equal to the net force, which is 8.785 N.

D) To find the maximum speed the object can have at the bottom before the string breaks, we need to find the maximum tension.

Given:
- Maximum tension (T_max) = 22.5 N
- Radius (r) = 0.850 m

Using the formula for tension:

Centripetal force (Tension) = (mass * velocity^2) / radius

Plugging in the values, we can rearrange the formula to solve for the maximum velocity:

T_max = (mass * velocity_max^2) / radius

Solving for velocity_max:

velocity_max^2 = (T_max * radius) / mass

velocity_max = √((T_max * radius) / mass)

Plugging in the values, we can calculate the maximum velocity:

velocity_max = √((22.5 * 0.850) / 0.275) = 13.18 m/s

Therefore, the maximum speed the object can have at the bottom before the string breaks is 13.18 m/s.

Remember, physics is no joke!

To answer these questions, we need to consider the forces acting on the object at different points along its path.

A) At any point along the path, there are two forces acting on the object:
1. Tension force (T): This force is exerted by the string and is directed towards the center of the circular path. It is responsible for keeping the object moving in a circular motion.
2. Weight force (mg): This is the force of gravity acting on the object, directed vertically downward.

B) Let's draw the free-body diagrams for the ball when it is at the bottom and top of the circle.

At the bottom:
In this position, the tension force (T) and the weight force (mg) are both present. Since the object is at the bottom, the weight force acts downwards, and the tension force acts upwards. Therefore, the free-body diagram will show the weight force (mg) pointing downward, and the tension force (T) pointing upward.

At the top:
In this position, the tension force (T) and the weight force (mg) are again present. However, this time, the tension force (T) acts downwards, and the weight force (mg) acts downwards as well. The free-body diagram will show both forces pointing downward.

C) To find the tension in the string at the top, we can use the centripetal force formula:
Fc = (mv^2) / r
Where:
- Fc is the centripetal force (which is provided by the tension)
- m is the mass of the object (0.275 kg)
- v is the speed of the object at the top (5.2 m/s)
- r is the radius of the circular path (0.85 m)

Plugging in the values, we get:
Fc = (0.275 kg * (5.2 m/s)^2) / 0.85 m

Solving this equation gives us the centripetal force, which is equal to the tension force at the top of the circle.

D) To find the maximum speed at the bottom before the string breaks, we need to again consider the tension force. Since the string breaks when its tension exceeds 22.5 N, we can set up an equation to find the maximum speed:
Tension = mv^2 / r

Rearranging the equation, we have:
v^2 = (Tension * r) / m

Plugging in the given values (Tension = 22.5 N, r = 0.85 m, m = 0.275 kg), we can solve for the maximum speed (v) at the bottom.

By following these steps, you should be able to answer all the questions and calculate the tension and maximum speed.

c) F = ma (in the 'y' direction)


*ball at top means 'T' and 'mg' are pointing "downward"*

-T -mg = ma
-T = mg + ma
T = -mg -ma
T = -m(g + v^2/r)

T = -.275(9.8 + (5.20^2/.85))

T = -11.4 N, or, 11.4 N downward