You must design a device for shooting a small marble vertically upward. The marb

Question

You must design a device for shooting a small marble vertically upward. The marble is in a small cup that is attached to the rim of a wheel of radius 0.230 m; the cup is covered by a lid. The wheel starts from rest and rotates about a horizontal axis that is perpendicular to the wheel at its center. After the wheel has turned through 20.0 rev, the cup is the same height as the center of the wheel. At this point in the motion, the lid opens and the marble travels vertically upward to a maximum height h above the center of the wheel. If the wheel rotates with a constant angular acceleration alpha, what value of alpha is required for the marble to reach a height of h = 12.0 m?

Explanation / Answer

Apply conservation of energy

initial,kinetic energy = final potentail energy

(1/2)*m*v^2 = m*g*h

v = sqrt(2*g*h)

= sqrt(2*9.8*12)

initial speed v = 15.34 m/s

required angular speed, w = v/r

W = 15.34/0.23 = 66.69 rad/s

angular dispalcement,

theta = 20 rev = 20*2*pi rad = 125.7 rad

let alpha ne the required angular acceleration,

So from the equation w^2 - wo^2 = 2*alpha*theta

alpha = (w^2 - wo^2)/(2*theta)

alpha = (66.69^2 - 0^2)/(2*125.7)

Angular accleration A = alpha= 17.69 rad/s^2

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