Calculating Average Torque on Coffee Can Down Ramp

Click For Summary
SUMMARY

The discussion focuses on calculating the average torque acting on a coffee can rolling down a ramp, utilizing the relationship between torque and angular acceleration defined by the equation τ = Iα. Key variables include the angle of incline, mass, radius of the coffee can, and the moment of inertia. The average torque can be determined using the formula τ_a = I_COM α, where α is derived from the linear acceleration of the center of mass (a_COM) calculated as a_COM = (g sin(θ)) / (1 + (I_COM / (MR^2))). Experimental methods to determine a_COM involve measuring the distance traveled down the ramp and the time taken.

PREREQUISITES
  • Understanding of rotational dynamics and torque
  • Familiarity with moment of inertia calculations
  • Knowledge of linear and angular acceleration concepts
  • Basic principles of kinematics in physics
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes, particularly cylinders
  • Learn how to experimentally measure angular acceleration using a rotational motion sensor
  • Explore the relationship between linear and angular motion through the use of the equations of motion
  • Investigate the effects of friction on rolling motion and torque calculations
USEFUL FOR

Students preparing for physics exams, particularly those focusing on mechanics, as well as educators and anyone interested in practical applications of rotational dynamics in experiments.

fiziksfun
Messages
77
Reaction score
0
1. I'm doing a lab with a coffee can rolling down a ramp. I'm asked to 'determine average torque' acting on the can while its rolling down the ramp. Then I'm asked to compare this value with I*alpha (angular acceleration) (I'm guessing because torque = I*alpha). So I need to calculate torque two ways, but I have no idea how!?

I know the angle of incline, the mass of the coffee can, the radius of the coffee can, the distance it was displaced in meters and radians, the angular and translation velocities (and accelerations) and its moment of inertia. Any ideas!?


2. Homework Equations



3. The Attempt at a Solution

I said

Fnet = Torque - Ff

this is just so horribly wrong.
 
Physics news on Phys.org
To calculate the average torque use

[tex]\tau _a = I_{COM} \alpha[/tex]

for the angular acceleration one have that

[tex]\alpha = \frac{a_{COM}}{R}[/tex]

the linear acceleration of the COM of the can down the ramp can be evaluated from

[tex]a_{COM} = \frac{g\sin(\theta)}{1 + \frac{I_{COM}}{MR^2}}[/tex]


The experimental determination of the average torque can be again be determined from the same equation by determining [tex]a_{COM}[/tex] experimentally

[tex]L = \frac{1}{2}a_{COM}t_L ^2[/tex]

where [tex]L[/tex] is the distance down the ramp and [tex]t_L[/tex] the time it takes to cover this distance. From which [tex]a_{COM}[/tex] can be calculated.
 
Last edited:
[tex] \theta = \omega_i + \frac{\alpha t^2}{2} = \frac{\alpha t^2}{2}[/tex]
[tex] 2\theta/t^2 = \alpha[/tex]
[tex] \theta = S_{ramp}/R_{can}[/tex]
[tex]\alpha = 2(S_{ramp}/R_{can}) / t^2[/tex][tex]\tau = I \alpha[/tex][tex]mgh - \frac{1}{2}mv^2 = \frac{1}{2}I \omega^2[/tex]
[tex]d = v_it + \frac{1}{2}at^2 = \frac{1}{2}at^2[/tex]
[tex]2d/t^2 = a[/tex]
[tex]v_f = v_i + at = 2d/t[/tex]

[tex]mgh - \frac{1}{2}m(2d/t)^2 = \frac{1}{2}I\omega^2[/tex]

[tex]\omega_f = \omega_i + \alpha t = \alpha t[/tex]
[tex]\omega_f = 2(S_{ramp}/R_{can}) / t[/tex]

[tex]2(2(S_{ramp}/R_{can}) / t)^2(mgh - \frac{1}{2}m(2d/t)^2)= I[/tex]

[tex]\tau = I \alpha[/tex]

[tex]2(2(S_{ramp}/R_{can}) / t)^2(mgh - \frac{1}{2}m(2d/t)^2) * 2(S_{ramp}/R_{can}) / t^2 = \tau[/tex]

Or something to that effect. This wasn't meant to learn from, it was meant to inspire. I'm sure I've made more than a few mistakes (point them out, I have an important physics exam in 2 days).
 

Similar threads

Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
4K
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K