Calculating Average Torque on Coffee Can Down Ramp

AI Thread Summary
To calculate the average torque on a coffee can rolling down a ramp, use the formula τ_a = I_COM * α, where α can be determined from linear acceleration a_COM = g * sin(θ) / (1 + I_COM / (MR^2)). The linear acceleration can be experimentally evaluated using the distance down the ramp and the time taken, with the equation L = 1/2 * a_COM * t_L^2. Additionally, the relationship between angular and linear motion can be utilized, such as θ = S_ramp / R_can and α = 2(S_ramp / R_can) / t^2. Accurate calculations are crucial for exam preparation, so reviewing the equations and ensuring correct application is recommended.
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

\tau _a = I_{COM} \alpha

for the angular acceleration one have that

\alpha = \frac{a_{COM}}{R}

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

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


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

L = \frac{1}{2}a_{COM}t_L ^2

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

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

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

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

\tau = I \alpha

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

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).
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top