Pulley, blocks, moment of inertia :S

[leanne1987]

pulley, blocks, moment of inertia... :S

Two blocks are connected by a light string passing over a pulley of radius 0.20m and moment of inertia I. The blocks move (towards the right) with an acceleration of 1.00m/s^2 along their frictionless inclines .

Mass A is on an incline of 32 degrees w.r.t the horizantal. and is 8 kg.
Mass B is on an incline of 61 degrees w.r.t the hoizantal, and is 10 kg.

Determine F (Ta), tension in the left part of the string.

Determine F (Tb), tensino in the right part of the string.

Find the net torque acting on the pulley. Take clockwise torques as positive.

Determine its moment of inertia, I/I].

*Hopefully I've done this right, just registered on this site a couple of minutes ago!

 

[Andrew Mason]

Do a free body diagram for each block. Write out an expression for the Tension acting on each block.

What is the net torque on the pulley (this is an expression in terms of Ta and Tb)?

What is the acceleration of this system?

Solve the equations for Ta and Tb.

AM

 

[kerimek]

Hello and welcome to our scientific community! Let's break down the given information and solve the problem step by step.

Firstly, we have two blocks (A and B) connected by a light string passing over a pulley. This setup allows us to use the principles of pulleys and moment of inertia to solve for the tension in the string and the net torque acting on the pulley.

To start, we need to find the acceleration of the blocks. Since the inclines are frictionless, we can use the formula a = gsinθ, where g is the acceleration due to gravity and θ is the angle of the incline. For block A, the angle is 32 degrees, so its acceleration is 8.3 m/s^2. For block B, the angle is 61 degrees, so its acceleration is 9.8 m/s^2.

Next, we can use Newton's second law (F = ma) to find the force acting on each block. For block A, the force is 66.4 N (8 kg x 8.3 m/s^2) and for block B, the force is 98 N (10 kg x 9.8 m/s^2).

Now, we can use the principles of pulleys to determine the tension in the string. Since the string is light, we can assume that its mass is negligible and the tension is the same throughout. This means that F(Ta) = F(Tb), so the tension in both parts of the string is 82.2 N (66.4 N + 98 N).

Moving on to the net torque acting on the pulley, we need to take into account the forces acting on it. The weight of block A creates a counterclockwise torque, while the weight of block B creates a clockwise torque. Since we are taking clockwise torques as positive, the net torque is 16.8 Nm (98 N x 0.20 m - 66.4 N x 0.20 m).

Finally, we can use the formula for the moment of inertia of a pulley (I = MR^2) to solve for its moment of inertia. Plugging in the values, we get a moment of inertia of 0.08 kgm^2.

I hope this helps and welcome again to our scientific community!