A 25 kilogram sphere is attached to the end of a robot arm. The robot arm is pro

Question

A 25 kilogram sphere is attached to the end of a robot arm. The robot arm is programmed, so the sphere potential energy is given by the following equation at t = 0, x = 0, so the sphere is initially at rest. The total mechanical energy of the system is 200 Joules.

g = 9.81 m/s2                U(x) = 6x2 - 10x                    E = 200 Joules

(a) Find the position of the sphere as a function of time.
(b) For t = 0.0147 seconds, find:
1.The velocity of the sphere.
2.The acceleration of the sphere.
3.The force acting on the sphere.

Note: Use the Metric system of units.

Explanation / Answer

m = 25 kg, x(0) = 0, v(0) = 0, E = 200 J, U(x) = 6x2 - 10x, g = 9.81 m/s2

(a) Find the position of the sphere as a function of time.

kinetice energy mv2/2 = E - U

v = [2(E - U)/m] = [2(200 - 6x2 + 10x)/25] = (16 - 0.48x2 + 0.8x) = dx/dt

dx/(16 - 0.48x2 + 0.8x) = dt

integrate, 5sin-1[(5 - 6x)/35]/23 = t + C

x(t) = {5 - 35sin[23(t + C)/5]}/6

use x(0) = 0, C = 5sin-1(1/7)/23 = 0.2069 rad

x(t) = {5 - 35sin[23(t + 0.2069)/5]}/6

(b) For t = 0.0147 seconds, find:
1.The velocity of the sphere.

x(0.0147) = 3.8739 m

v(0.0147) = (16 - 0.48x2 + 0.8x) = 3.449 m/s
2.The acceleration of the sphere.

a = dv/dt = dv/dx * dx/dt = dv/dx * v = (1/2) d(v2)/dx = (1/2) d(16 - 0.48x2 + 0.8x)/dx = -0.48x + 0.4

= -0.48*3.8739 + 0.4 = -1.46 m/s2
3.The force acting on the sphere.

F = ma = -36.5 N

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