An amusement park game, shown in (Figure 1) , launches a marble toward a small c

Question

An amusement park game, shown in (Figure 1) , launches a marble toward a small cup. The marble is placed directly on top of a spring-loaded wheel and held with a clamp. When released, the wheel spins around clockwise at constant angular acceleration, opening the clamp and releasing the marble after making 1112revolution. The top of the cup is level with the center of the wheel.

What angular acceleration is needed for the ball to land in the cup?

Explanation / Answer

let the marble is taking time t to reach the cup

Apply, -h = -0.5*g*t^2

==> t = sqrt(2*h/g)

= sqrt(2*0.4/9.8)

= 0.2857 s

speed of the barble when it leaves the wheel, v = ditsnace travelled/time taken

= 1/0.2857

= 3.5 m/s

angular speed of the wheel whe it leaves the marble leaves the wheeel,

w = v/r

= 3.5/0.4

= 8.75 rad/s

initial angular speed, wo = 0

angular displacement, theta = 1112 rev

= 1112*2*pi

= 6987 revolutions

Let alfa is the angular acceleration.

Apply, w^2 - wo^2 = 2*alfa*theta)

alfa = (w2^2 - wo^2)/(2*theta)

= (8.75^2 - 0^2)/(2*6987)

= 5.48*10^-3 rad/s^2 <<<<<<-------Answer

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