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I have the question "An Lp record rotates at 33 and a third rpm(revolutions per minute). What speed does the record rotate under the needle when the needle is 14cm from the centre hole?"

So I first made the radius 14 cm into 0.14m, And the rpm from 33 and a third into 0.6 s.

I then found the period T = 1/f = 1/0.6s = 1.7s.

I then used the equation V = wr to find the speed.

w = 2pi/1.7s = 3.7 rads^-1.

Therefore, V = (3.7 rads^-1) x (0.14 m)

Therefore V = 0.51[ms^-1]

Is this correct ?

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Dan
  • 1,649

1 Answers1

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The thing $33$ and a third is $100/3$. Note that frequency=$100/3$ per minute or $5/9$ per second, radius=$0.14m$.

Since it is asking for speed, you just apply the given data in formula for speed in such kind of rotation which is $V=2\pi rv$ where $v$ is frequency.

Applying all the given data in formula, you get that$V=2\pi rv=2\times \frac{22}{7}\times0.14\times\frac{5}{9}$. Solve and get the answer

Vidyanshu Mishra
  • 10,143