A rod suspended on its end and acting as a physical pendulum swings with a period of 0.7 s. What is its length? (g = 9.80 m/s^2)
2006-11-07 15:03:27 · 4 answers · asked by tonberry79 3 in Science & Mathematics ➔ Physics
T=2pi*sqrt(L/g) for pedulum with center of mass at the end, so rod length is twice that, so T=2pi*sqrt(L_rod/2/g), the answer is L=g*T^2/2/pi^2=0.243m.
The length is not 12.2cm because the center of mass of the rod is half distance of the length of the rod
2006-11-07 15:58:00 · answer #1 · answered by justiceforall 2 · 0⤊ 0⤋
T=2pirt(l/g)
squaring
T^2=4pi^2l/g
l=gT^2/(4pi^2)
=9.8*(0.7)^2/4(3.14)^2
=0.122 m
=12.2 cm
2006-11-08 01:57:33 · answer #2 · answered by raj 7 · 0⤊ 0⤋
.7=2(pi)* sq of L/9.8
2006-11-07 23:41:46 · answer #3 · answered by Matt C 1 · 0⤊ 0⤋
L=4 * 3.1415 * 3.1415 * T * 9.8
L=4 * 3.1415 * 3.1415 *.7 *9.8
2006-11-08 04:09:59 · answer #4 · answered by sona 1 · 0⤊ 0⤋