Inertial Disk Lab: Physics 11 H Experiment

[agentd00nut]

Ok, in my Physics 11 H class we did a lab with an inertial disk (a disk with a hub in the center, a string is attached to the hub like a huge half a yoyo where the yoyo is stationary and weights are the fingers falling down) and i need to be able to calculate the tension in the string for various weights attatched to the end of the strings. I am pretty sure that just using T=ma will work or maybve do something with tension but i don't know where the radius of the hub comes into play and the weight of the disk, any help would be greatly appreciated the diamter of the hub is 76 cm and the disk is 25 cm i calculated the acceleration of a 69.9 gram weight to be .039 m/s^2 and it falls 1.75 meters in a 6.69 second period so using my logic t= .669 * .039 is that correct?

 

[lightgrav]

let's see, if it starts from v=0 and moves 1.75 m in 6.69 s,
its avg velocity is 0.2616 m/s .
the final v = .5312 m/s , same as its change of velocity.
the acceleration is 0.0782 m/s/s . Where's your .039 from?

the mass (0.0669 kg !) is pulled down by gravity, up by Tension.
Most places use g = 9.81 N/kg unless you know yours better.
The SUM of Forces cause the mass to accelerate.

 

[quicknote]

Hi there,

Thank you for sharing your experiment with us! It sounds like you are working on calculating the tension in a string attached to an inertial disk with various weights attached to the end. You are correct that the equation T=ma can be used to calculate the tension, but there are a few other factors that you will need to consider.

First, the radius of the hub will play a role in determining the tension. The tension in the string will vary depending on where the weight is attached on the string. For example, if the weight is attached closer to the hub, the tension will be greater than if it is attached further away. This is because the weight will be closer to the center of rotation, and therefore will require less force to maintain the circular motion.

Second, the weight of the disk itself will also affect the tension in the string. The more mass the disk has, the greater the force required to keep it in circular motion, and therefore the greater the tension in the string.

To calculate the tension, you will need to use the equation T=ma, where T is the tension, m is the mass of the weight attached to the string, and a is the acceleration of the weight. However, you will also need to take into account the radius of the hub and the mass of the disk. Here is an example of how you could calculate the tension for a 69.9 gram weight attached to the end of the string, using the information you provided:

First, we need to calculate the mass of the entire system (the weight, the string, and the disk). Let's say the string has a mass of 10 grams and the disk has a mass of 200 grams. Then the total mass would be 69.9 + 10 + 200 = 279.9 grams.

Next, we need to calculate the acceleration of the weight. You have already calculated this to be 0.039 m/s^2.

Now, we can plug these values into the equation T=ma to calculate the tension. T = (279.9 grams) * (0.039 m/s^2) = 10.92 grams*m/s^2, or approximately 0.11 N.

This is just an example, and the actual tension will vary depending on the specific values of the radius and the mass of the disk. I would recommend doing a few more trials with different weights and positions on the string to get a better