A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of Us = 0.80 and Uk = 0.50. The turntable very slowly speeds up to 60 rpm. Does the coin slide off?
How do you work through this??? Thanks!
How do you work through this??? Thanks!
-
The maximum centripetal force required
to hold coin in a 0.15 m radius circular path at 60 RPM is:
Fc =mV²/R
{m = 0.005 kg, R = 0.015 m, V = wR = (60/60)2π(0.015) = 2π(0.015) = 0.0942 rad/s}
Fc = (0.005)(0.0942)²/(0.015) = 2.96E-3 N
the force of STATIC friction = 0.80(mg) = (0.80)(0.005)(9.81) = 3.92E-2 N
3.92E-2 > 2.96E-3 so coin remains fixed (doesn't slide off turntable) ANS
to hold coin in a 0.15 m radius circular path at 60 RPM is:
Fc =mV²/R
{m = 0.005 kg, R = 0.015 m, V = wR = (60/60)2π(0.015) = 2π(0.015) = 0.0942 rad/s}
Fc = (0.005)(0.0942)²/(0.015) = 2.96E-3 N
the force of STATIC friction = 0.80(mg) = (0.80)(0.005)(9.81) = 3.92E-2 N
3.92E-2 > 2.96E-3 so coin remains fixed (doesn't slide off turntable) ANS