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A small ball of mass 0.75 kg is attached to one end of a 1.25-m-long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is 30o from the vertical,what is the magnitude of the gravitational torque calculated about the pivot

Sagot :

leena

Hi there!

Recall the equation for Torque:
[tex]\large\boxed{\Sigma \tau = rFsin\phi}[/tex]

In words, the torque is the force multiplied by the distance from the LINE OF ACTION to the PIVOT POINT.

In this instance:

F = Mg = 0.75 · 9.8 = 7.35 N

r = 1.25 m

sin(30) = 0.5

Plug in the values:

[tex]\Sigma \tau = (1.25)(7.35)(0.5) = \boxed{4.59 Nm}[/tex]

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