Solving Masses Connected by Rope: Acceleration & Forces

[khawar]

A 4 kg mass is connected by a light cord to a 3 kg mass on a smooth surface.The cord is around a frictionless axle and has a moment of inertia of .5 kg*m^2 and a radius of .3m. Assuming that the cord does not slip arount the axle,

a) what is the acceleration of the two masses

b)what are the forces of tension in the rope that connects to the two masses

I began this problem using the principle that if, as the first question shows, there is acceleration, then torque net is not equal to zero. In such a situation, the following equation is recommended:

moment of inertia* angular acceleration= sum of the torques. The equation for torque is force*distance from axis.

Apparently in such a problem, the torque net equation above must be supplemented with force net equations for each of the two masses. The force net equation is:

Force net *linear acceleration= sum of forces

By substituting the unknowns in the force net equation into the torque net equation, one should be able to find the acceleration and then the two forces.

 

[Nate810]

a) The acceleration of the two masses is given by the equation: 0.5 * angular acceleration = (3 * 9.81 * 0.3) + (4 * 9.81 * 0.3) Solving for the angular acceleration, we get: angular acceleration = (3 * 9.81 * 0.3) + (4 * 9.81 * 0.3) / 0.5 Therefore, the angular acceleration = 14.73 m/s^2. The linear acceleration of each mass can be found using the equation: linear acceleration = angular acceleration * radius Therefore, the linear acceleration of the 3 kg mass is 4.42 m/s^2 and the linear acceleration of the 4 kg mass is 5.82 m/s^2. b) The forces of tension in the rope that connects to the two masses can be found using the equation: Force net *linear acceleration = sum of forces Therefore, the force of tension between the 3 kg mass and the 4 kg mass is (3 * 4.42) + (4 * 5.82) = 37.64 N.

 

[M98Ranger]

First, let's set up the coordinate system. Let the positive x-axis be pointing to the right, and the positive y-axis be pointing upwards. The axle is located at the origin (0,0). The 4 kg mass is located at (0.3,0), and the 3 kg mass is located at (0.6,0).

a) To find the acceleration, we can use the force net equation for each mass. For the 4 kg mass, the force net equation is:

F - T = ma

Where F is the force applied to the mass, T is the tension in the rope, m is the mass, and a is the acceleration. We can rearrange this equation to solve for a:

a = (F-T)/m

Similarly, for the 3 kg mass, the force net equation is:

-T = ma

Again, we can rearrange this equation to solve for a:

a = -T/m

Since both masses are connected by the same rope, the tension in the rope will be the same for both masses. This means we can set the two acceleration equations equal to each other and solve for T:

(F-T)/4 = -T/3

Solving for T, we get:

T = 3F/7

Now, we can substitute this value of T into either of the acceleration equations to find the acceleration of the masses. Let's use the first one:

a = (F-3F/7)/4

Simplifying, we get:

a = F/28

So, the acceleration of the two masses is directly proportional to the applied force F.

b) To find the forces of tension in the rope, we can use the value of T that we found in part a. The force of tension in the rope connecting the 4 kg mass is:

T = 3F/7

And the force of tension in the rope connecting the 3 kg mass is:

T = 3F/7

So, the forces of tension in the rope are equal and opposite, as expected in a system where the rope does not slip around the axle.