Another Angular Acceleration Problem

AI Thread Summary
The discussion revolves around calculating the torque exerted on a CD that accelerates uniformly from rest to 450 rev/min over 3.0 revolutions. The initial calculations using the formula tau = I * alpha were attempted, but the user encountered errors in determining alpha. It was clarified that the correct approach does not involve taking the square root in the equation for alpha, which should be alpha = wf^2/(2*theta). Despite correcting this, the user still found discrepancies in their final answer compared to the expected torque value of 0.0072 N*m. The conversation highlights confusion over the calculations and the possibility of an incorrect provided answer.
jbgibson
Messages
33
Reaction score
0
The problem states: when the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius of 6.0-cm and a mass of 17-g, what is the torque exerted on it?

I used the the forumla tau = I * alpha; I = .5(0.017)(0.06)^2 = 3.06E-5;
in order to calculate alpha, I used wf^2 = wo^2+2alpha(theta); from this I get alpha = square root (wf^2)/(2*theta). I realize some converting is necessary so I convert 450 rev/min into 47.124 rad/s, and 3.0 revolutions into 18.50 rad. Applying these figures to solve for alpha, I get 60.018 rad/s^2. Finally, tau = I*alpha = 0.00184 N*m.

This is incorrect. Can someone direct me in the right direction? Thanks in advance.
 
Physics news on Phys.org
You have two equations, omega = alpha * t, and fi = ½ * alpha * t^2, these equations are like v = at and s = ½at^2 in linear motion. You also have two unkwons, so you can solve both.

Then use M = J*alpha.
 
alpha = wf^2/(2 theta) , no sqrt.
you see Work and Energy hiding in this kinematic equation?
 
lightgrav said:
alpha = wf^2/(2 theta) , no sqrt.
you see Work and Energy hiding in this kinematic equation?

I still don't have a clue where to go with this. I was in error, there is no sqrt., but the solution I'm getting is incorrect. The answer to this problem is 0.0072-N*m. How is that? What am I doing wrong?
 
I get the same answer as you. Maybe the given answer is incorrect.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top