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

A)

for inclined plane ,

Static friction = mg sin theta where theta is angle with horizontal

here theta = 90 - 80 =10 degree, m =10*0.4535 kg = 4.535 kg

Static friction = mg sin 10o = 10*0.4535 kg*9.8 m/s2 *Sin 10 degree = 7.717 N

B) Let it is moving with angular velocity w

mg cos10o - N = mw2 sin10o

N = mg cos10o - mw2 sin10o *0.25cos 10o= 4.535*0.985 - 4.535*0.174 W2 *0.246 = 4.47 - 0.194w2

mg sin 10 - mw2 cos10 r= uN =0.3*( 4.47 - 0.194w2)

7.717 - 4.535*0.985*0.246w2 = 1.341- 0.058w2

1.098w2 -0.058w2 =7.717-1.341

1.04 w2 =6.376

w2 = 6.13

w=2.476 /s

revolution per minute= w/2pi =60* 2.476/6.28 = 23.66 revs/min

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