Calculating Rotational Inertia with Blocks and a Pulley

[dudforreal]

In Figure, block 1 has mass m1 = 430 g, block 2 has mass m2 = 540 g, and the pulley is on a frictionless horizontal axle and has radius R = 5.3 cm. When released from rest, block 2 falls 80 cm in 5.3 s (without the cord slipping on the pulley). What is the pulley's rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution.

The Attempt at a Solution

I tried using the T-mg=ma for both the small mass and the large mass to get T and substitute the T values in (T2-T1)r=Iα to get I, the rotational inertia, but ended up getting the wrong answer. I don't know what is wrong but my acceleration for block 2 was very insignificant because I ended up with 0.028301886.

 

[Doc Al]

I don't know what is wrong but my acceleration for block 2 was very insignificant because I ended up with 0.028301886.

How did you calculate the acceleration?

 

[dudforreal]

first i calculated the velocity using distance divided by time and got 0.15 and then i used v=u+at with u=o and t=5.3

 

[Doc Al]

first i calculated the velocity using distance divided by time and got 0.15 and then i used v=u+at with u=o and t=5.3

Distance divided by time will give you the average velocity. Better to find another kinematic formula, one that directly relates distance, time, and acceleration.

 

[dudforreal]

i do not recall any other equations that can help but is v=u+at good enough?

 

[Doc Al]

i do not recall any other equations that can help but is v=u+at good enough?

That equation is fine, if you use the correct velocity. In that equation, v is not the average velocity.

See: https://www.physicsforums.com/showpost.php?p=905663&postcount=2"

 

[dudforreal]

how do you use that to get rotational motion?

 

[Doc Al]

how do you use that to get rotational motion?

Knowing the acceleration will eventually allow you to find the rope tensions and the rotational inertia of the pulley.

Show the equations you wrote for each block and for the pulley. You mentioned them in your first post.

 

[dudforreal]

yeah i still get the same value for acceleration

 

[Doc Al]

yeah i still get the same value for acceleration

What did you do differently?

 

[dudforreal]

i basically used the kinematics equations and came up with the same answer

 

[Doc Al]

i basically used the kinematics equations and came up with the same answer

What specific kinematic equations did you use?

 

[dudforreal]

v=u+at, v^2-u^2=2as, s=ut+1/2at^2

 

[Doc Al]

s=ut+1/2at^2

That's the one I'd use.

 

[dudforreal]

what values did u put in?

 

[Doc Al]

what values did u put in?

You tell me what you put in. (Realize that it starts from rest.)

 

[dudforreal]

well i would use u=o, t=5.3 and s=0.8 is this correct?

 

[Doc Al]

well i would use u=o, t=5.3 and s=0.8 is this correct?

Yep. That's what I would use.

 

[dudforreal]

i still got a very small acceleration of 0.057m/s^2

 

[Doc Al]

i still got a very small acceleration of 0.057m/s^2

Looks fine to me. Yes, it's small, but so what?

 

[dudforreal]

what do i do afterwards?

 

[dudforreal]

how do i get the acceleration of the smaller mass and alpha in the rotational inertia equation?

 

[Doc Al]

what do i do afterwards?

The acceleration is just data. You'll plug it into your equations to solve for the rotational inertia.

You need to apply Newton's 2nd law to each body: the two blocks and the pulley. (For the pulley, you'll use torque instead of force.) You started this in your first post.

You'll get three equations. (One for each body.) You'll solve them together to find the unknowns, one of which will be the rotational inertia of the pulley.

 

[dudforreal]

i can't find the acceleration of the smaller mass in the data

 

[Doc Al]

i can't find the acceleration of the smaller mass in the data

The masses are connected by the rope. They have the same acceleration. (The same magnitude of acceleration; the direction is different for each.)