Translational/Rotation Question (Pulley with a Mass)

[dan38]

1. Homework Statement
Compound pulley with a mass (10kg) attached to outer drum
R = Outer Radius (250 mm)
r = Inner Radius (150 mm)

Horizontal force of 250 N being applied at edge of inner drum
Friction torque of 7 N m present
Mass of pulley = 20 kg and it has radius of gyration of 160 mm

2. Homework Equations



3. The Attempt at a Solution

I * Angular Acceleration + 10* Acceleration * R = F*r - 7

Where I = mk^2 of the pulley
and Acceleration = Angular Acceleration * R

Basically used these equations and inputted the given values, but still got the wrong answer.

 

[Achy47]

Hello, based on the information provided, I believe there may be a few errors in your equations. Here is my attempt at a solution:

1. First, let's determine the net torque acting on the pulley. The only forces acting on the pulley are the applied force of 250 N and the friction torque of 7 Nm. This gives us a net torque of 250 N * (150 mm) - 7 Nm = 368 Nm.

2. Next, we can use the net torque to find the angular acceleration of the pulley. We know that the moment of inertia (I) of the pulley is given by I = mk^2, where m is the mass of the pulley and k is the radius of gyration. Plugging in the given values, we get I = (20 kg) * (0.16 m)^2 = 0.512 kgm^2. Now, we can use the equation τ = Iα to find the angular acceleration, α = τ / I = 368 Nm / 0.512 kgm^2 = 718.75 rad/s^2.

3. Finally, we can use the angular acceleration to find the linear acceleration of the mass attached to the outer drum. We know that the linear acceleration (a) is related to the angular acceleration (α) by the equation a = α * R, where R is the radius of the pulley. Plugging in the given values, we get a = (718.75 rad/s^2) * (0.25 m) = 179.69 m/s^2.

I hope this helps and please let me know if you have any further questions or if you would like me to clarify any steps in my solution.