A playground carrousel is a disk with radius 1.5m. The children push it, making

Question

A playground carrousel is a disk with radius 1.5m. The children push it, making it turn at a rate of 4.0rpm, they can stop pushing and the disk slows down at a constant rate, coming to a stop in 15.0 seconds.

a) What is the angular accelration?
I got 0.279 Rad/s^2 but my teacher got 0.0279 rad/s^2

b) At the moment 12 seconds after the children stopped pushing what is the magnitude of the total acceleratio of a point on the edge of the carrousel?

c)How many revolutions odes the carousel make during the 15 seconds it is slowing to a stop?

d) A child leaves his lunch box on the outer edge of the carrousel. What maximum constant angular speed at which the carrousel could rotate with out the lunch box flying off? The coefficient of kinetic friction between the lunchbox and the carrousel is .270.

Explanation / Answer

a) 4.0 rpm = 2 (4.0) / 60s = i ; f = 0 ; r = 1.5m ; t = 15s ; = ?

Assuming constant deceleration, f = i + t ; = (f - i) / t = (0 - 2 (4.0) / 60s ) / 15s = - 0.0279 rev/s2

b) because it's constant acceleration, the deceleration is the same. So I'll assume you're asking for angular velocity.

f = i + t using this formula again, just change f into ? and t = 12s

f = 2 (4.0) / 60s +  - 0.0279 rev/s2 (12s) = 0.0840 rad/s

c) f = i + it + 1/2t2 so f = 0 + 2 (4.0) / 60s (15s) + 1/2 (- 0.0279 rev/s2) (15s)2 = 3.14 rads = 0.5 rev

d) again, I'll assume you meant static friction because if it's kinetic friction, the lunchbox is moving already and will eventually slide off.

frictional force ƒ = F = mg ; the maximum of which will equal the centrifugal force which is equal to the centripetal force. so mg = m v2/r ; g = v2/r ; v = rg ; v = r ; rg = r ; = rg / r

= {.270 (1.5m) (9.8m/s2)} / 1.5m = 1.33 rad/s = 12.683 rpm

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