Answer:
62.64 RPM.
Explanation:
Given that
m= 4.6 g
r= 19 cm
μs = 0.820
μk = 0.440.
The angular speed of the turntable = ω rad/s
Condition just before the slipping starts
The maximum value of the static friction force =Centripetal force
[tex]\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s[/tex]
[tex]\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM[/tex]
Therefore the speed in RPM will be 62.64 RPM.
