Finding Angular Velocity in Rotational Motion Problems

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To find angular velocity in rotational motion problems, converting 53 rpm to radians per second yields 5.55 rad/sec, which can be further multiplied by 2π to get approximately 34.87 rad/sec. The discussion emphasizes that this is primarily a conversion problem, and mass is not relevant to the calculation of angular velocity. Participants highlight the importance of understanding unit conversions and the specific requirements of the problem. There is a concern about the accuracy of the final answer, indicating potential confusion in the calculations. Overall, the focus is on mastering conversion techniques for solving similar rotational motion problems.
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Homework Statement
A ball, 1.8 kg, is attached to the end of a rope and spun in a horizontal circle above a student's head. As the student rotates the ball in a horizontal clockwise circle their lab partner counts 53 rotations in one minute. What is the ball's angular velocity in radians per second?
Relevant Equations
1 rad/sec = 60/2pi rmp
53 rpm equals 5.55 rad/sec
multiply 5.55 by 2pi to get angular velocity of 34.8717

Is the answer 34.8717?

What should I have done to more accurately solve the problem with a better understanding?

What other steps should I take when solving similar problems?

and lastly,
Is the mass relevant to the problem in any way?
 
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momoneedsphysicshelp said:
53 rpm equals 5.55 rad/sec
Right. Why continue past that? You are asked for it in units of rad/sec.
 
momoneedsphysicshelp said:
Is the mass relevant to the problem in any way?
No.
 
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haruspex said:
Right. Why continue past that? You are asked for it in units of rad/sec.
So this is only a conversion problem?
Thanks you very much.
 
When I turned in that answer, it was still wrong.
 
What answer ?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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