10kg mass is attached to a 5kg mass by a 20cm cord. The 5 kg mass is attached by a 30cm cord to a motor that rotates the entire mass/cord system 5rev/s in the horizontal plane. What is the tension in each cord?

To find the tension in each cord, we can use Newton's second law and consider the forces acting on each mass separately.

Let's start with the 10kg mass. The tension in the cord connected to the 10kg mass will provide the centripetal force required to keep the mass moving in a circular path. The centripetal force can be obtained using the formula:

F_c = m * a

where F_c is the centripetal force, m is the mass, and a is the acceleration.

The acceleration can be calculated using the relationship between linear velocity and angular velocity:

v = ω * r

where v is the linear velocity, ω is the angular velocity, and r is the radius (length of the cord in this case).

Since the angular velocity ω is given as 5 rev/s and the radius r is 20cm (0.2m), we can calculate the linear velocity:

v = 5 rev/s * 2π * 0.2m = 10π m/s

Now we can calculate the centripetal acceleration:

a = v^2 / r = (10π)^2 / 0.2 ≈ 157.91 m/s^2

Substituting the known values, we can find the tension in the cord attached to the 10kg mass:

F_c = 10kg * 157.91 m/s^2 = 1579.1 N

Therefore, the tension in the cord attached to the 10kg mass is approximately 1579.1 Newtons.

Now let's consider the 5kg mass. This mass experiences two forces: the tension in the cord pulling it outward and the gravitational force pulling it downward. The net force acting on the 5kg mass should provide the acceleration required to keep it in the circular path.

The net external force can be calculated using the formula:

F_net = m * a

Since the only external force acting on the 5kg mass is the tension in the cord, we can write:

F_net = Tension - Weight

where Tension is the tension in the cord and Weight is the gravitational force acting on the mass:

Weight = m * g

where m is the mass and g is the acceleration due to gravity.

Substituting the known values, we can calculate the weight of the 5kg mass:

Weight = 5kg * 9.8 m/s^2 ≈ 49 N

Since the mass is not accelerating vertically (it is in equilibrium in this direction), the net vertical force is zero:

F_net = 0

Now we can write the equation for the net external force:

0 = Tension - 49 N

Solving for Tension, we find:

Tension = 49 N

Therefore, the tension in the cord attached to the 5kg mass is 49 Newtons.

In summary, the tension in the cord attached to the 10kg mass is approximately 1579.1 Newtons, and the tension in the cord attached to the 5kg mass is 49 Newtons.

To find the tension in each cord, we can use Newton's second law and consider the forces acting on each mass separately.

Let's start with the 10 kg mass:

1. Find the acceleration of the system:
- The system is rotating at 5 revolutions per second.
- One revolution is equal to 2π radians.
- Therefore, the angular velocity (ω) is 5 * 2π = 10π rad/s.
- The tangential velocity (v) of the 10 kg mass is given by v = ω * r, where r is the radius of the circular path.
- The radius can be found using the cord length, which is 20 cm or 0.2 m.
- Hence, r = 0.2 m.
- Therefore, v = 10π * 0.2 = 2π m/s.
- Since the system is rotating with constant velocity, the acceleration is zero (because it is moving in a circle).

2. Calculate the sum of forces acting on the 10 kg mass:
- There are two forces acting on the 10 kg mass: tension in the cord and the gravitational force.
- The gravitational force is given by F = m * g, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s^2).
- Hence, F_gravity = 10 kg * 9.8 m/s^2 = 98 N (downward).
- The tension force is directed towards the center of the circular path and provides the necessary centripetal force.
- Since the acceleration is zero, the tension force has to cancel out the gravitational force.
- Therefore, the tension in the cord attached to the 10 kg mass is equal to the gravitational force: T_10kg = 98 N (upward).

Next, let's move on to the 5 kg mass:

3. Find the acceleration of the 5 kg mass:
- The 5 kg mass is moving in a circular path of radius 30 cm or 0.3 m.
- The velocity (v_5kg) of the 5 kg mass can be calculated using the angular velocity:
v_5kg = ω * r_5kg = (10π rad/s) * (0.3 m) = 3π m/s.
- Since the system is rotating with constant velocity, the acceleration of the 5 kg mass is also zero.

4. Calculate the sum of forces acting on the 5 kg mass:
- Similar to the 10 kg mass, there are two forces acting on the 5 kg mass: tension in the cord and the gravitational force.
- The gravitational force is F_gravity_5kg = 5 kg * 9.8 m/s^2 = 49 N (downward).
- The tension in the cord attached to the 5 kg mass provides the centripetal force required for circular motion.
- Since the acceleration is zero, the tension force has to cancel out the gravitational force.
- Therefore, the tension in the cord attached to the 5 kg mass is equal to the gravitational force: T_5kg = 49 N (upward).

So, the tension in the cord attached to the 10 kg mass is 98 N (upward), and the tension in the cord attached to the 5 kg mass is 49 N (upward).