Question

A small ball of mass 0.64 kg is attached to one end of a 0.840-m-long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is 26° from the vertical, what is the magnitude of the torque about the pivot?

Answer 1

Answer:

The torque about the pivot is

T=2.3 N*m

Explanation:

$m=0.64kg\\L=0.84m\\\alpha=26$

The gravitational torque calculated about the pivot is giving by:

T=abs(Fg)*L*sin(α)

$Fg=m*g\\Fg=0.64kg*9.8\frac{m}{s^{2} } \\Fg=6.272 N$

Remplacing:

$T=Fg*L*sin(\alpha)\\T=6.272*0.84m*sin(26)\\T=5.26848*0.4383\\T=2.3N*m$