How Do You Calculate the Acceleration of a Bucket on a Pulley System?

[dolpho]

Homework Statement

A 2.85 kg bucket is attached to a disk-shaped pulley of radius .121 m and mass .742 kg. If the bucket is allowed to fall, what is the linear acceleration, angular acceleration, and how far does it drop in 1.5 seconds.

I really only need help with the first part since the other two are pretty easy and depend on the first answer.

Homework Equations

t - mg = ma
alpha = a / r

The Attempt at a Solution

I'm a bit confused on which signs I should be using. This was my first try and I'd love some input to see where I went wrong.

T - mg = ma
Since we don't know or a we need to use the torque produced by the pulley.
torque = TR = I alpha
T= .5Mr^2 alpha / R
alpha = a / R
So then T = .5Ma ----> we plug this into T in the original equation

.5Ma - mg = ma,

.5(.742)(a) - (2.85)(9.8) = 2.85a
.371a - 27.93 = 2.85a

-27.93 = 2.85a - .371a
= -27.93 = 2.479a
a = - 11.26

Now I'm pretty sure I messed up on the signs somewhere. Wouldn't it all depend on how you set up the diagrams? So in this case the tension is pointing upwards so it's positive, and MG is - so it's negative. Also, how would you know if the pulley is going clockwise or counter clockwise?

Would love any clarification of how to set up the signs. Thanks guys!

 

[haruspex]

I'm pretty sure I messed up on the signs somewhere.

It shouldn't matter what you assume about the directions things will move in provided you're consistent.
According to your first equation, T - mg = ma, you're measuring acceleration as upwards. The equation alpha = a / R then implies alpha is in the direction of reeling in the bucket. But the string effectively reverses the direction of T; if it's upwards on the bucket then it's downwards on the pulley. So for the third equation you would have T= -.5MR² alpha / R.

 

[tiny-tim]

hi dolpho!

T - mg = ma
Since we don't know or a we need to use the torque produced by the pulley.
torque = TR = I alpha
T= .5Mr^2 alpha / R
alpha = a / R
So then T = .5Ma ----> we plug this into T in the original equation

.5Ma - mg = ma,
…
Now I'm pretty sure I messed up on the signs somewhere. Wouldn't it all depend on how you set up the diagrams? So in this case the tension is pointing upwards so it's positive, and MG is - so it's negative. Also, how would you know if the pulley is going clockwise or counter clockwise?

that's fine down to the last line, which should be .5Ma - mg = -ma

you assumed that α was positive in the direction of the string (ie downwards), so T had the same sign as α

but if α is positive downwards, then the bucket is falling, so a (of the bucket) must also be positive downwards

(of course, you could have assumed α was positive upwards, then a would be also, and they would both come out as having negative values!)

[btw, there's a simple way of checking your result (but which i don't think would be approved of in the exam) …

use the "rolling mass", I/r² (= m/2 for a cylinder), so the total weight is Mg, and the total effective mass is M + m/2 ]​

 

[dolpho]

Great, thanks for the help. I'll definitely try that check :D

 

[Nickie]

Your approach to solving this problem is correct. However, there are a few errors in your calculations. First, when you calculated the torque, you forgot to include the weight of the bucket (2.85 kg) in the calculation. The correct equation should be T = (0.5)(0.742 + 2.85)(a) = 1.796a.

Secondly, when you substituted this value for T in the original equation, you also forgot to include the weight of the bucket in the equation. The correct equation should be 1.796a - (2.85)(9.8) = 2.85a.

Now, to address your question about the signs, it all depends on how you define your coordinate system. In this problem, the tension in the rope is acting upwards, so it would have a positive sign. The weight of the bucket is acting downwards, so it would have a negative sign. As for the direction of rotation of the pulley, it does not affect the calculations in this problem. However, if it is necessary to determine the direction, you can use the right-hand rule, where the direction of the fingers curling around the pulley in the direction of rotation will give the direction of the angular velocity.

As for the linear and angular acceleration, they can be determined using the equations you provided. The linear acceleration can be found by solving for a in the equation a = (1.796a - 27.93) / 2.85 = -9.2 m/s^2. The angular acceleration can be found by substituting this value for a in the equation alpha = a / r = (-9.2) / 0.121 = -75.8 rad/s^2.

Finally, to determine how far the bucket drops in 1.5 seconds, you can use the equation s = ut + (1/2)at^2, where u is the initial velocity (which is 0 m/s in this case). Substituting the values, we get s = (1/2)(-9.2)(1.5)^2 = -10.4 m. Therefore, the bucket drops 10.4 meters in 1.5 seconds.

Remember to always double check your calculations and include all relevant values in your equations. Keep up the good work!