A standard cassette tape is placed in a standard cassette player. Each side play

Question

A standard cassette tape is placed in a standard cassette player. Each side plays for 45 minutes. The two tape wheels of the cassette have a diameter of 2.0 cm when fully wound and fit onto two 1.0 cm diameter spindles in the player. Suppose that a motor drives one spindle at constant angular speed of 0.90 rad/s and the other spindle is free to rotate at any angular speed. Find the tape's thickness.

I keep coming up with the wrong answer. I did 0.90 * 45 * 60/(2p)=386.75. Then I divided 1.0/2.0=0.5; divided that by the 386.75 to get 1.29*10^-3. This answer is incorrect and I can't figure out what I'm doing wrong?

Please help!!

Explanation / Answer

when the tape wounded over the spindle Assume tape the wounded over thick ness be t = 2cm (because cassette has a diameter 2 cm ) If the No. of revolutions through which spindle rotated in 45 min   is N Angular displacement = N . 2 radians (since : d = d /d t ) t = N . 2 N = t / 2     = (0.9 rad/s )(45 x 60 sec) / 2 ( rad/ rev ) N = 386.9 rev i.e 2 cm width is divided in to 386.9 turns of tape thickness then each turn thickness : t = 2 cm / 386.9      = 5.169 x 10-3  cm Assume tape the wounded over thick ness be t = 2cm (because cassette has a diameter 2 cm ) If the No. of revolutions through which spindle rotated in 45 min   is N Angular displacement = N . 2 radians (since : d = d /d t ) t = N . 2 N = t / 2     = (0.9 rad/s )(45 x 60 sec) / 2 ( rad/ rev ) N = 386.9 rev i.e 2 cm width is divided in to 386.9 turns of tape thickness then each turn thickness : t = 2 cm / 386.9      = 5.169 x 10-3  cm

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