A DJ is playing an Armin van Buuren record on Turntable A at a typical 33.3 rpm,

Question

A DJ is playing an Armin van Buuren record on Turntable A at a typical 33.3 rpm, while a Blank & Jones record rests motionless on Turntable B. At a given time, he turns on one record and turns off the other. If the Armin van Buuren record rotationally decelerates at one-fifth the rate of rotational acceleration of the Blank & Jones record, at what rotational speed will the two turntables match when they finally "synch"? ______rpm

If this process took 2.53 seconds, how many revolutions would the Armin van Buuren record sweep? ______rev

Explanation / Answer

The DJ is playing at an angular velocity=33.3 rpm

It is given that the acceleration is 5 times the deceleration

So B accelerates to 5/6 of the speed while A decelerates 1/6 of the speed .

So B's speed= 33.3 rpm*5/6=27.65 rpm

Therefore, they match at 5/6(33.3) rpm.

The average rotational speed of record A during this interval is 11/12 of 33.3 rpm

average of 1 x 33.3 and 5/6 x 33.3 = ((1 + 5/6)/2) x 33.3.)=30.525

. Divide that number by 60 to get the number of revolutions per second

So number of revolutions per second=0.508

In 2.53 seconds the number of revolutions is =0.508*2.53=1.287 rev

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.