A 1.00kg , horizontal, uniform tray is attached to a vertical ideal spring of fo

Question

A 1.00kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 190N/m and a 270g metal ball is in the tray. The spring is below the tray, so it can oscillate up-and-down. The tray is then pushed down 12.2cm below its equilibrium point (call this point A) and released from rest.

A) How high above point A will the tray be when the metal ball leaves the tray? (Hint: This does notoccur when the ball and tray reach their maximum speeds.)

B) How much time elapses between releasing the system at point A and the ball leaving the tray?

C) How much time elapses between releasing the system at point A and the ball leaving the tray?

Explanation / Answer

w = (k / m)^1/2 = (190 / 1.27)^1/2 = 12.2 / sec       angular frequency
x = A cos w t         at t = 0       x = -.122 = A
x = -.122 cos 12.2 t
v = -w * A sin w t = -12.2 * -.122 sin 12.2 t = 1.49 sin 12.2 t     
a = - w^2 A cos w t = - (12.2^2 * -.122) cos 12.2 t = 18.2 cos 12.2 t
For ball to leave pan   a = -g = -9.8
cos 12.2 t = -9.8 / 18.2 = - .538
12.2 t = 123 deg = 2.14 rad
t = 2.14 / 12.2 = .175 sec        time at which ball leaves tray
x = -.122 cos 123 =  .066 m = 6.6 cm

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