Solve Rotation Questions: 2.53 Revs in 6.28 & 19.3 Secs

  • Thread starter Thread starter shell4987
  • Start date Start date
  • Tags Tags
    Rotation
AI Thread Summary
The discussion revolves around solving a physics problem involving a merry-go-round's rotation with an angular acceleration of 1.65 rad/s². The initial attempts to calculate the time for the first and second 2.53 revolutions were incorrect due to unit conversion errors, specifically not converting revolutions to radians. The correct approach involves using the equation relating angular displacement, initial angular velocity, time, and angular acceleration. After correcting the unit conversions, the revised calculations yield approximately 1.75 seconds for the first 2.53 revolutions and 2.48 seconds for the next 2.53 revolutions. Accurate unit consistency is crucial for solving rotation problems effectively.
shell4987
Messages
43
Reaction score
0

Homework Statement


A merry-go-round rotates from rest with an angular acceleration of 1.65 rad/s2. How long does it take to rotate through (a) the first 2.53 rev and (b) the next 2.53 rev?


Homework Equations





The Attempt at a Solution


I tried to do this and I converted the rev to radians by multiplying the revolutions by 2pi and then for (a) I divided 15.896/1.65 and got 6.28 seconds (15.896 is 2.53(2pi))and for (b) I divided 31.79/1.65 to get 19.3 seconds. What am I doing wrong here?
 
Physics news on Phys.org
I divided 15.896/1.65 and got 6.28 seconds
Check your units, this doesn't work out to seconds.

What equation relates time, angular displacement, and angular acceleration?
 
hage567 said:
Check your units, this doesn't work out to seconds.

What equation relates time, angular displacement, and angular acceleration?

theta-theta initial= w initial(t) + (1/2)at^2?
 
What do you get for the time with that equation?
 
hage567 said:
What do you get for the time with that equation?

For (a) I got 1.75seconds and for (b) I got 2.48 seconds. I used 2.53 as the first displacement and for w I put in 0 and for acceleration I put 1.65 and then solved for t and that's what I got, and for the second part I put in 5.06 for the displacement. What am I doing wrong?
 
I used 2.53 as the first displacement

You need to change this into radians before using it in your calculations. Your units are not consistent otherwise. Try part (a) again with that change.
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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