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

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

Step-by-step 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* F\\\\\tau=(45-28)* 10^(-2)* 0.3* 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.