A 56.5 g box is left 11.2 cm from the centre of a merry-go-round. If the box has coefficients of friction of μs = 0.720 and μk = 0.520 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?
Question
A 56.5 g box is left 11.2 cm from the centre of a merry-go-round. If the box has coefficients of friction of μs = 0.720 and μk = 0.520 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?
Solution
To solve this problem, we need to use the concept of centripetal force and frictional force.
Step 1: Convert the mass of the box from grams to kilograms and the distance from the center of the merry-go-round from cm to meters.
m = 56.5 g = 0.0565 kg
r = 11.2 cm = 0.112 m
Step 2: The maximum static frictional force (fs_max) that can act on the box before it starts to slide is given by fs_max = μs * m * g, where g is the acceleration due to gravity (9.8 m/s²).
fs_max = 0.720 * 0.0565 kg * 9.8 m/s² = 0.39984 N
Step 3: This frictional force provides the necessary centripetal force to keep the box moving in a circle. The centripetal force (Fc) required to keep an object moving in a circle of radius r at a speed v is given by Fc = m * v² / r.
0.39984 N = 0.0565 kg * v² / 0.112 m
Step 4: Solve the above equation for v.
v = sqrt(0.39984 N * 0.112 m / 0.0565 kg) = 2.38 m/s
So, the maximum speed of the merry-go-round without having the box slide off is approximately 2.38 m/s.
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