Pulley system with Moment of Inertia

[TrippingBilly]

A block of mass m1 = 2.05 kg and a block of mass m2 = 5.95 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. These blocks are allowed to move on a fixed block-wedge of angle = 30.0°. The coefficient of kinetic friction is 0.360 for both blocks. Draw free-body diagrams of both blocks and of the pulley. (The mass m2 is sitting on a 30 degree slope with the horizontal, which is connected by a string to the pulley at the top of the slope, and m1 is on the other side of the pulley on a flat surface.)

Before I can start to work on this problem, I'm not sure what the equation for the moment of inertia of the pulley would be. I don't know what direction to head in once I figure that but I can give it a try before I come back for more help.

 

[Nate810]

M1: (m1)(g)sin(30°) + (m1)(a) - (m1)(μk)(g)cos(30°) = 0M2: (m2)(g)sin(30°) + (m2)(a) - (m2)(μk)(g)cos(30°) = 0Pulley:I = MR^2F_tension * R = I * alpha

 

[kyle8921]

The moment of inertia of a solid disk is given by the equation I = (1/2)MR^2, where M is the mass of the disk and R is the radius. In this case, the mass of the pulley is given as 10.0 kg and the radius is 0.250 m. Therefore, the moment of inertia of the pulley is I = (1/2)(10.0 kg)(0.250 m)^2 = 0.313 kgm^2.

Now, let's draw the free-body diagrams for each block and the pulley. For block m1, there are three forces acting on it: its weight (mg), the tension in the string (T), and the normal force from the surface it is on (N). The normal force is perpendicular to the surface and the weight is acting straight down. The tension in the string is in the direction of motion, which is to the left. Therefore, the free-body diagram for m1 would look like this:

|N
|
|____T <--- direction of motion
|
|mg

For block m2, there are also three forces acting on it: its weight (mg), the tension in the string (T), and the normal force from the surface it is on (N). However, since the block is on a slope, the normal force is not directly perpendicular to the surface. Instead, it can be broken down into two components: one perpendicular to the surface and one parallel to the surface. The weight is still acting straight down, and the tension in the string is still in the direction of motion, which is down the slope. Therefore, the free-body diagram for m2 would look like this:

|N
|\
| \____T <--- direction of motion
| \
| mg

For the pulley, there are two forces acting on it: the tension in the string (T) and the normal force from the surface it is on (N). The normal force is perpendicular to the surface and the tension in the string is in the direction of motion, which is down. Since the pulley is rotating, there is also a torque acting on it. The torque is given by the equation τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. In this case, the torque is caused