A 88.6 g drink is left on a turntable causing it to turn with an angular speed of 8.80 rad/s. If the drink has coefficients of friction of μs = 0.970 and μk = 0.470 with the turntable, what is the furthest distance that the drink could have been placed from the centre of the turntable for the drink not to slide? 0.0125 m 0.0595 m 0.00135 m 0.123 m
Question
A 88.6 g drink is left on a turntable causing it to turn with an angular speed of 8.80 rad/s. If the drink has coefficients of friction of μs = 0.970 and μk = 0.470 with the turntable, what is the furthest distance that the drink could have been placed from the centre of the turntable for the drink not to slide? 0.0125 m 0.0595 m 0.00135 m 0.123 m
Solution
To solve this problem, we need to use the formula for the maximum static friction force, which is the force that prevents the drink from sliding. This force is given by:
F_max = μs * N
where μs is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the weight of the drink, which is its mass times the acceleration due to gravity (g = 9.8 m/s^2). So we have:
N = m * g = 88.6 g * 9.8 m/s^2 = 0.0866 kg * 9.8 m/s^2 = 0.84828 N
Then, the maximum static friction force is:
F_max = μs * N = 0.970 * 0.84828 N = 0.82289 N
This force must be equal to the centripetal force, which is the force that keeps the drink moving in a circle. The centripetal force is given by:
F_c = m * ω^2 * r
where m is the mass of the drink, ω is the angular speed, and r is the distance from the center of the turntable. We can set these two forces equal to each other and solve for r:
F_max = F_c 0.82289 N = 0.0866 kg * (8.80 rad/s)^2 * r 0.82289 N = 6.65632 kgm/s^2 * r r = 0.82289 N / 6.65632 kgm/s^2 = 0.123 m
So, the furthest distance that the drink could have been placed from the centre of the turntable for the drink not to slide is 0.123 m.
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