Please do number 1 parts A through E (no need to do number 2)and thank you! Home
ID: 1558086 • Letter: P
Question
Please do number 1 parts A through E (no need to do number 2)and thank you!
Homework Homework Name Physics 101 1. A simple pendulum consists of a small ball of mass m 200g swinging on a massless, inextensible string of length R -7m. At time f 0 the pendulum is released from the horizontal position, as shown below. A short time later, the pendulum reaches the bottom of its swing and encounters a nail sticking out of the page a distance L 6m below the pivot point. The pendulum begins to wrap around the nail, and at time t it is at the top of its first rotation -L a nail A. Is mechanical energy of the earth-ball system conserved during the interval from t 0 to t top Explain your reasoning. B. Find the speed of the ball at time t op (Assume g Show your work. C. Find the magnitude and direction of the acceleration of the ball at E op Show your work. HW-1Explanation / Answer
a. the mechanical energy of the earth-ball system is conserved
As the ball moves from horizontal to 6 meters below, it's kinetic energy increases as its potential energy decreases. If we neglect the force of air resistance, the increase of kinetic energy is equal to the decrease of potential energy.
conserve energy
change in KE = change in PE
0.5mv^2 = mgh
0.5 * 0.2 * v^2 = 12
v = 10.95 m/s
When the string hits the nail, the nail exerts a force the string. At this position, one meter of the string is below the nail. Since the ball has kinetic energy, it will continue moving. When the string hits the nail, one meter of the string is below the nail.
PE gained by ball after ithits the nail
Increase of PE = 0.2 * 10 * 1 = 2 J
12 – 2 = 10 J
This is the kinetic energy of the ball, when it is one meter above its lowest position. Let’s determine the velocity of
the ball at this position.
0.5 * 0.2 * v^2 = 10
v = 10 m/s
The force which the nail exerted on the string did not affect the total mechanical energy of the earth-ball system. So, the mechanical energy of the earth-ball system is conserved.
b. v = 10 m/s
c.
Centripetal acceleration = v^2/r
The radius is 1 meter
Centripetal acceleration = 10^2/1 = 100 m/s^2
Since the ball is above the center of the circle, the direction of the acceleration is down!
d.
Since the ball is moving in a circle, the net force is the centripetal force
Fc = m* v^2/r = 0.2 * 10^2/1 = 20 N
e.
At the top, the weight of the ball and the tension force are pointing toward the center the circle.
Weight + Tension = Fc
0.2 * 10 + T = 20
T = 20 – 2 = 18 N