object of mass 2 kg is attached to string and rotate in vertical circle of radius 1 m. at highest point it has velocity of 4 m/s. what is tension in string at that point. at lowest point it is .5 m above ground. if string breaks at highest point how far does it travel horizontally before it hits ground

To find the tension in the string at the highest point, we can use the concept of centripetal force and gravitational force. At the highest point, the tension in the string should be equal to the sum of the centripetal force and gravitational force acting on the object.

1. Centripetal force (Fc):

The centripetal force acting on the object is given by the formula:
Fc = mv^2 / r

Where:
m = mass of the object = 2 kg
v = velocity of the object = 4 m/s
r = radius of the circle = 1 m

Plugging in the values, we get:
Fc = (2 kg)(4 m/s)^2 / 1 m
Fc = 32 N

2. Gravitational force (Fg):

The gravitational force acting on the object is given by the formula:
Fg = mg

Where:
m = mass of the object = 2 kg
g = acceleration due to gravity = 9.8 m/s^2

Plugging in the values, we get:
Fg = (2 kg)(9.8 m/s^2)
Fg = 19.6 N

3. Tension in the string (T):

At the highest point, the tension in the string can be calculated by adding the centripetal force and gravitational force:
T = Fc + Fg
T = 32 N + 19.6 N
T = 51.6 N

Therefore, the tension in the string at the highest point is 51.6 N.

Now, let's calculate how far the object travels horizontally before hitting the ground if the string breaks at the highest point.

Since the object is 0.5 m above the ground at the lowest point, we can calculate the time it takes to reach that point using the formula:

h = (1/2) * g * t^2

Where:
h = distance traveled vertically = 0.5 m
g = acceleration due to gravity = 9.8 m/s^2
t = time taken to reach the lowest point

Rearranging the equation, we get:

t = sqrt(2h / g)
t = sqrt(2 * 0.5 m / 9.8 m/s^2)
t = sqrt(0.102 m / 9.8 m/s^2)
t = sqrt(0.0104 s^2)
t ≈ 0.102 s

Now, we can calculate the horizontal distance traveled using the formula:

d = v * t

Where:
v = horizontal velocity of the object at the lowest point (which is the same as the horizontal component of the velocity at the highest point)
t = time taken to reach the lowest point

Given that the string broke at the highest point, the horizontal velocity at that point is equal to the velocity at the highest point (4 m/s). Therefore:

d = (4 m/s) * (0.102 s)
d ≈ 0.408 m

Hence, the object will travel approximately 0.408 meters horizontally before hitting the ground if the string breaks at the highest point.

To find the tension in the string at the highest point of the vertical circle, we need to consider the forces acting on the object.

At the highest point, the object is moving in a circle, so its acceleration is directed towards the center of the circle. For the object to move in a circle, the sum of the forces acting on it must provide the centripetal force.

The net force acting on the object at the highest point is the tension in the string minus the gravitational force:

T - mg = (mv^2) / r

where T is the tension in the string, m is the mass of the object, g is the acceleration due to gravity, v is the velocity of the object, and r is the radius of the circle.

Given that the mass of the object is 2 kg, the radius of the circle is 1 m, and the velocity of the object at the highest point is 4 m/s, we can substitute these values into the equation:

T - (2 kg)(9.8 m/s^2) = (2 kg)(4 m/s)^2 / 1 m

T - 19.6 N = 32 N

Simplifying the equation, we find:

T = 32 N + 19.6 N

T = 51.6 N

Therefore, the tension in the string at the highest point is 51.6 N.

Now, let's consider the second part of the question, where the string breaks at the highest point and we need to determine how far the object travels horizontally before it hits the ground.

When the string breaks, the object will follow a projectile motion. We can find the time it takes for the object to hit the ground by using the fact that the height at the lowest point is 0.5 m.

We can use the equation for vertical motion:

d = vi * t + (1/2) * a * t^2

where d is the distance (which is 0.5 m), vi is the initial vertical velocity (which is 0 m/s as the object is momentarily at rest at the highest point), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight.

Substituting the values:

0.5 m = 0 m/s * t + (1/2) * (-9.8 m/s^2) * t^2

Simplifying the equation:

-4.9 * t^2 = 0.5

t^2 = 0.5 / 4.9

t = sqrt(0.102)

t ≈ 0.319 s

Now, we can find the horizontal distance traveled by the object using the equation:

d = vt

where d is the horizontal distance, v is the horizontal velocity, and t is the time of flight.

The horizontal velocity of the object remains constant throughout the motion. We can find the horizontal velocity using the velocity at the highest point, which is 4 m/s. Since there are no horizontal forces acting on the object, the horizontal velocity remains constant.

Therefore, the horizontal distance traveled is:

d = (4 m/s) * (0.319 s)

d ≈ 1.276 m

Hence, if the string breaks at the highest point, the object will travel approximately 1.276 meters horizontally before hitting the ground.