Question

The center of mass of a 0.30-kg (non-uniform) meter stick is located at its 45-cm mark. What is the magnitude of the torque (in N⋅m) due to gravity if it is supported at the 28-cm mark? (Use g = 9.79 m/s2). (NEVER include units with the answer to ANY numerical question.)

Answer 1

Answer:

The magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.

Explanation:

Given that,

Mass of the meter stick, m = 0.3 kg

Center of mass is located at its 45 cm mark.

We need to find the magnitude of the torque due to gravity if it is supported at the 28-cm mark. Torque acting on the object is given by :

$\tau=r\times F\\\\\tau=(45-28)\times 10^{-2}\times 0.3\times 9.79\\\\\tau=0.499\ N-m$

or

$\tau=0.5\ N-m$

So, the magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.