A simple pendulum with a small, 7.0 kg lead ball is released from rest at an ang

Question

A simple pendulum with a small, 7.0 kg lead ball is released from rest at an angle of 30.0 as shown. A horizontal spring is connected to an immovable block as shown. Sometime after the pendulum ball collides with the spring the speed of the ball is zero. Using the Principle of Conservation of Energy, find theta. About how fast was the ball traveling just before it encountered the spring? You may assume that theta is small so that cos(theta)~1 and sin(theta) ~ theta. Ignore the mass of the pendulum rope and that of the spring. The spring has a force constant of 500 N/m. Show all work. Hint: Be sure to include the energy of the ball and the spring at all points in the motion.

Explanation / Answer

to find the velocity just befire the encounter we apply cons, of energy to get

= 1/2 mV^2 +P.E at the instant = 1/2mV1^2 + 0

p.e of the ball =m*g*[ l-lcos[30] ]= 7*9.8*3[1-.866] =27.44 J

now we get back to our equation

at the initial state K,E = 0

so 27.44J = 0.5*7*V1^2

V1=2.8 m/s

now to find the value of theta we again apply cons of energy

= 27.44J=1/2k[lsin{THETA}]^2 + MGL[1-cos THETA]

by solving this equation we get the value of theta

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