2. A pyramid is at the center of a merry-go-round. The face of the pyramid is 80

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)

Explanation / Answer

here,

angle of pyramid with vertical, A = 80

weight of block, w = 10 lb = 44.482 N

mass of block, m = 44.482/g = 44.482/9.81 = 4.534 kg

slope distance, d = 25 cm = 0.25 m

Part A: Since Merry go Round is not moving so Centripitl force will be Zero,
Ff = us*mg ( us coefficient of stativ friction)
Ff = 0.3 * 4.354 * 9.81 * Cos80
Ff = 2.225 N

Part B:
When Merry go round start rotating,
Fc - Ff = 0 ( From Newton Second Law, SUM(F) = 0 )
FF = Fc
uk*mg*Cos80 = mv^2/r

Since V = w*r ( w is angular velocity)

so,
uk*mg*Cos80 = m*w^2*r^2/r

solving for angular velocity, w

w = sqrt((uk*g*Cos80)/r)

w = sqrt((0.15*9.81*Cos80)/0.25)

w = 1.011 rad/s

1 rad/s = 9.55 rev/min
so 1.011 rad/s = 1.011*9.55 = 9.654 rev/min

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