2. A pyramid is at the center of a merry-go-round. The face of the pyramid is 80
ID: 1439394 • Letter: 2
Question
2. A pyramid is at the center of a merry-go-round. The face of the pyramid is 80 degrees to vertical. A block that weighs 10 lbs sits on one of the sides of the pyramid, at a sloped distance of 25.0 cm (from the center of the pyramid to the center of the block). The coefficients of static and kinetic friction are 0.300 and 0.150, respectively.
A. The merry-go-round is not moving. What is the static friction force in Newtons?
B. The merry-go-round starts turning, accelerating very slowly. How fast is it turning when the block slips? (In revs/minute)
a) The static friction force is simply (Coefficient of static)(Fnormal). In this case the static friction force is 13.3N. Please use this in part B.
Explanation / Answer
A) static friction force = Coefficient of static * N
= 0.300 * 4.536 * 9.8 * cos10
= 13.3 N
B) Here, mgsin(theta) - F = mv2* cos(theta)/r
=> 4.536 * 9.8 * sin10 - 6.65 = 4.536 * v2 * cos10/0.246
=> v = 0.2426 m/sec
=> `turning speed = 0.2426/0.246
= 0.9861 rad/sec
= (0.9861*60)/(2*3.14) rev/min
= 9.421 revs/minute
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