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 ?
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