Two circus clowns (each having mass of 50kg) swing on two flying trapezes (negligible mass, length 25m). At peak of the swing, one grabs the other, and the two swing back to one platform. The time for the forward and return motion in s is?
2007-02-01 11:41:45 · 1 answers · asked by Shane W 1 in Science & Mathematics ➔ Physics
Sounds like a pendulum problem to me.
I heard that a period of the pendulum depends on the length of the ‘arm’ and the local gravitational acceleration.
Like in T=2 pi sqrt(l/g)
T- period
Pi – 3.14...
sqrt – square root ;-)
l – lengt of the arm or the distance from the pivotal point to the center of the mass.
g – gravitational acceleration usually 9.81 m/s^2
t(total)= T1 + T2
T1= time it takes for them to swing at the peak and for them to one to grab the other
T2= time to swing back to the platform.
Since mass play no role on the period and the swing did not elongate and hopefully g remained the same we have
t(total)=2T=4 pi sqrt(l/g)
t(total)=4x3.14 x sqrt(25/9.81)=20sec
2007-02-02 01:19:06 · answer #1 · answered by Edward 7 · 0⤊ 0⤋