Disk speed NON UNIFORM and UNIFOR CIRCULAR MOTION

[sevens]

A computer disk is 8.0cm in diameter. a refrence dot on the edge of the disk is initially located at theta= 45 degrees.

The disk accelerates steadily for 1/2 second, reaching 2000 rpm, then coasts at steady angular velocity for 1/2 second.

what is the speed of the refrence dot at 1 second. answer is in m/s.

it seems to me that the first half second of this problem could be ignored, since after the first 1/2 second we stop accelerating and are in Uniform Circular Motion.

so I thought i would use v = ( 2(pi)r ) / T.
i knew my radius was .04m.

and to find T i did ( 2000 rev) * (1 min / 60 sec) = 33.33 rev/sec

my V ended being .0075 m/s this answer is wrong and I am not sure if it is because i ignored the first half second of the problem, i just don't see how that ties into finding the velocity at 1 second. i figured that if it was coasting steadily at 2000 RPM it turns into a unifor circular motion problem



any help is apreciated, thanks.

 

[Doc Al]

it seems to me that the first half second of this problem could be ignored, since after the first 1/2 second we stop accelerating and are in Uniform Circular Motion.

Good.

so I thought i would use v = ( 2(pi)r ) / T.
i knew my radius was .04m.

Good.

and to find T i did ( 2000 rev) * (1 min / 60 sec) = 33.33 rev/sec

T is the period, the time for one revolution (sec/rev). You calculated the inverse of the period (rev/sec).

 

[ZapperZ]

A computer disk is 8.0cm in diameter. a refrence dot on the edge of the disk is initially located at theta= 45 degrees.

The disk accelerates steadily for 1/2 second, reaching 2000 rpm, then coasts at steady angular velocity for 1/2 second.

what is the speed of the refrence dot at 1 second. answer is in m/s.

it seems to me that the first half second of this problem could be ignored, since after the first 1/2 second we stop accelerating and are in Uniform Circular Motion.

so I thought i would use v = ( 2(pi)r ) / T.
i knew my radius was .04m.

and to find T i did ( 2000 rev) * (1 min / 60 sec) = 33.33 rev/sec

my V ended being .0075 m/s this answer is wrong and I am not sure if it is because i ignored the first half second of the problem, i just don't see how that ties into finding the velocity at 1 second. i figured that if it was coasting steadily at 2000 RPM it turns into a unifor circular motion problem



any help is apreciated, thanks.

You have a bit of a problem with dimensional analysis here. First of all, you're working way too hard! Secondly, T is the PERIOD of revolution, i.e. it has the dimension of time. If you look above, you have T having units of rev/sec. This is a frequency, not a period.

Start with the fact that you know that

v = r \omega

This is one of those problem in which you could solve it if you know what each of these symbols mean PHYSICALLY. Here, \omega means angular velocity or angular frequency, and it has units of radians/second. If an object makes one completely rotation in 1 second, then it has an angular velocity of 2 \pi/sec.

Now look at yourproblem. When it is coasting with constant angular velocity, it is rotating at 2000 rpm, or rotations per minute. It means that in one minute, it is making 2000*2pi radians of rotations. This is the angular velocity \omega! If you convert the minute into seconds, you have now the angular velocity in rad/sec, just what you need to solve this problem.

Notice that what I did above made very little use of "formulas". All I did was make use of the definitions of each of the quantities involved and extract the values I need from those definitions.

Zz.

 

[sevens]

Thanks for the help

what i did was take my answer of 33.33 rev/sec and did (1/33.33rev/sec) to find the amount of time per rotation or period T. giving me .03 sec. now that i knew the right T i could solve my equation

v= (2(pi)r)/T = (2(pi)(.04m))/.03 sec = 8.38 m/s

thanks for all the help.