A 1.00 kg mass is attached to a string wrapped around a cylinder of radius 0.05m. Two masses are connected with a rod and attached to the end of the cylinder, perpendicular to its axis. as the mass falls from rest the rotating contraption spins through 12 pi radians in 1.2 seconds.

A) Find the angular acceleration of the rotating object.
B) Using the falling mass, find the tension in the string.
C) Calculate the rotational inertia of the object.

12 pi = (1/2) alpha (1.2)^2

so solve for alpha

a = acceleration down of mass = r alpha
= .05 alpha
so
1 g - T = 1 a
T = 1(g-a)

Torque = .05 T = I alpha
solve for I

A) To find the angular acceleration (α), we can use the formula:

α = Δθ / Δt

where Δθ is the change in angle (12π radians) and Δt is the change in time (1.2 seconds).

Substituting the given values:

α = (12π radians) / (1.2 seconds)
= 10π radians per second squared

Therefore, the angular acceleration of the rotating object is 10π radians per second squared.

B) To find the tension in the string, we need to consider the forces acting on the falling mass. The only force acting on it is the tension in the string.

Using Newton's second law, we can write the equation:

F = m * a

where F is the force, m is the mass, and a is the acceleration.

In this case, the force is the tension in the string, and the acceleration is the acceleration due to gravity (g).

T = m * g

Given that the mass is 1.00 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can substitute the values:

T = (1.00 kg) * (9.8 m/s^2)
T = 9.8 N

Therefore, the tension in the string is 9.8 Newtons.

C) To calculate the rotational inertia of the object (I), we need to consider the moment of inertia of the cylinder and the masses connected to it.

The moment of inertia (I) of a cylinder is given by the formula:

I = 0.5 * m * r^2

where m is the mass of the cylinder and r is the radius of the cylinder.

In this case, the mass of the cylinder is unknown, so we can solve for it using the given information.

The length of the cylinder is not provided, so we cannot directly use the given mass of 1.00 kg to calculate the mass of the cylinder. However, we can use the information that the rotating contraption spins through 12π radians in 1.2 seconds to find the angular speed (ω).

ω = Δθ / Δt

Substituting the given values:

ω = (12π radians) / (1.2 seconds)
= 10π radians per second

The moment of inertia (I) of the cylinder is related to the angular speed (ω) and angular acceleration (α) by the formula:

I = (m * r^2) / (α)

Substituting the known values:

I = (m * (0.05m)^2) / (10π radians per second^2)
= (m * 0.0025m^2) / (10π radians per second^2)
= (0.00025m^2 * m) / (10π radians per second^2)
= 0.000025m^2 / (10π radians per second^2)
= 0.0000025m^2 per radian per second^2

Therefore, the rotational inertia of the object is 0.0000025 kg m^2 per radian per second^2.

To find the answers to these questions, we can start by understanding the relevant concepts.

A) Angular acceleration (α) is the rate of change of angular velocity (ω). It is given by the formula α = (Δω / Δt), where Δω is the change in angular velocity and Δt is the change in time. We are given that the rotating object spins through 12π radians in 1.2 seconds, so we can find the angular velocity (ω) using the formula ω = Δθ / Δt, where Δθ is the change in angle.

B) The tension in the string can be found using Newton's second law of motion. In this case, we will consider the forces acting on the falling mass. The two forces are the weight (mg) and the tension in the string (T). Since the mass is in free fall, the net force acting on it is the tension in the string, given by T = mg.

C) The rotational inertia (or moment of inertia) of the object measures its resistance to rotational motion and depends on the mass distribution and geometry of the object. The rotational inertia (I) can be calculated using the formula I = MR^2, where M is the mass of the rotating object and R is the radius of the cylinder.

Now, let's calculate the answers to the questions:

A) We are given that the rotating object spins through 12π radians in 1.2 seconds. The angular velocity (ω) is Δθ / Δt, which is (12π radians) / (1.2 seconds) = 10π rad/s. The angular acceleration α can be calculated as α = (Δω / Δt), which is (10π rad/s) / (1.2 seconds) = 8.33 rad/s^2.

B) The tension in the string can be found using Newton's second law of motion. Since the mass is in free fall, the net force acting on it is the tension in the string (T), given by T = mg. The mass is 1.00 kg, and the acceleration due to gravity (g) is approximately 9.8 m/s^2. Therefore, T = (1.00 kg) * (9.8 m/s^2) = 9.8 N.

C) The rotational inertia (I) can be calculated using the formula I = MR^2, where M is the mass of the rotating object and R is the radius of the cylinder. The mass is 1.00 kg. The radius (R) is given as 0.05 m. Therefore, I = (1.00 kg) * (0.05 m)^2 = 0.0025 kg·m^2.

So the answers to the questions are:
A) The angular acceleration of the rotating object is 8.33 rad/s^2.
B) The tension in the string is 9.8 N.
C) The rotational inertia of the object is 0.0025 kg·m^2.