A merry-go-round of radius R rotates at constant speed with period T. What minimum Co-effi-cient of static friction between merry-go-round and a box mass, m, placed at its edge the willenable the box to remain on the surface without sliding?
Question
A merry-go-round of radius R rotates at constant speed with period T. What minimum Co-effi-cient of static friction between merry-go-round and a box mass, m, placed at its edge the willenable the box to remain on the surface without sliding?
Solution
To solve this problem, we need to find the minimum coefficient of static friction (μ) that will prevent the box from sliding off the merry-go-round.
Step 1: Identify the forces acting on the box. The box is subjected to three forces:
- The gravitational force (mg) acting downwards.
- The normal force (N) acting upwards.
- The frictional force (f) acting towards the center of the merry-go-round.
Step 2: Apply Newton's second law in the radial direction. The net force acting on the box in the radial direction is the frictional force, which provides the necessary centripetal force for the box to move in a circle. So, we have:
f = m * (v^2 / R)
where v is the speed of the box, which can be expressed in terms of the period T as v = 2πR / T.
Step 3: Substitute v into the equation. f = m * ((2πR / T)^2 / R) f = 4π^2mR / T^2
Step 4: The frictional force is also equal to the product of the normal force and the coefficient of static friction. f = μN
Step 5: The normal force is equal to the weight of the box. N = mg
Step 6: Substitute N into the equation. f = μmg
Step 7: Set the two expressions for the frictional force equal to each other and solve for μ. 4π^2mR / T^2 = μmg μ = 4π^2R / (gT^2)
So, the minimum coefficient of static friction that will prevent the box from sliding off the merry-go-round is μ = 4π^2R / (gT^2).
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