A 1.00-m long meter-stick of mass 0.125 kg is pined at one end and can rotate in
ID: 1512423 • Letter: A
Question
A 1.00-m long meter-stick of mass 0.125 kg is pined at one end and can rotate in the Vertical plane. The meter stick is initially kept horizontal. the meter-stick is then released. Answer the following question for when the meter stick reaches the vertical position (I_rod = mL^2/3) a. What is the change in the gravitational potential energy of the meter stick? b. Find the kinetic energy and the angular speed of the meter stick when reaches vertical position. c. What is the speed of the free end of the meter stick at the bottom?Explanation / Answer
Ans:-
a)change in PE = final - initial
final is at bottom so = 0
initial = mgh
therefore:
0 - mgh = -(0.125)(9.8)(0.5)
= - 0.613J
b) Work = KE + PE
here work is 0 (no friction)
KE = final - intial
final = 1/2Iw^2
initial = 0
in this case I = 1/3ML^2
therefore:
1/6ML^2w^2 = 0.613
1/6(0.125)(1^2)w^2 = 0.613
w = sqrt(0.613*6/0.125)
w = 5.4244rad/s
KE =1/6*ML^2w^2 =1/6*0.125*1^2*5.4244^2 = 0.613J
c) w = V/R
R = 1m
therefore:
w = V
5.4244m/s = V
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.