A 75.0 g box is left 17.9 cm from the centre of a merry-go-round. If the box has coefficients of friction of μs = 0.830 and μk = 0.490 with the merry-go-round, what is the maximum speed of the merry-go-round without having the box slide off of the merry-go-round? 49.5 rpm 20.6 rpm 6.74 rpm 64.4 rpm
Question
A 75.0 g box is left 17.9 cm from the centre of a merry-go-round. If the box has coefficients of friction of μs = 0.830 and μk = 0.490 with the merry-go-round, what is the maximum speed of the merry-go-round without having the box slide off of the merry-go-round? 49.5 rpm 20.6 rpm 6.74 rpm 64.4 rpm
Solution
To solve this problem, we need to use the formula for the maximum static friction force, which is F_max = μs * m * g, where μs is the coefficient of static friction, m is the mass of the box, and g is the acceleration due to gravity.
Step 1: Convert the mass of the box from grams to kilograms. 75.0 g = 0.075 kg
Step 2: Calculate the maximum static friction force. F_max = μs * m * g = 0.830 * 0.075 kg * 9.8 m/s^2 = 0.60945 N
This force is what keeps the box from sliding off the merry-go-round. It must be equal to the centripetal force, which is m * v^2 / r, where v is the speed of the merry-go-round and r is the distance from the center of the merry-go-round to the box.
Step 3: Set the maximum static friction force equal to the centripetal force and solve for v. 0.60945 N = 0.075 kg * v^2 / 0.179 m v = sqrt(0.60945 N * 0.179 m / 0.075 kg) = 1.42 m/s
Step 4: Convert the speed from m/s to rpm. 1 m/s = 60/2π rpm = 9.55 rpm v = 1.42 m/s * 9.55 rpm = 13.56 rpm
So, the maximum speed of the merry-go-round without having the box slide off is approximately 13.56 rpm. This is not one of the options given, so there may be a mistake in the problem or in the given options.
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