Question

The center of mass of a 0.30 kg (non-uniform) meter stick is located at its 48-cm mark. What is the magnitude of the torque (in N⋅m) due to gravity if it is supported at the 25-cm mark? (Use g = 9.79 m/s2). (NEVER include units with the answer to ANY numerical question.)

Answer 1

`\tau=0.675\ Nm`

Step-by-step explanation:

It is given that,

Mass of the meter stick, m = 0.3 kg

Meter stick is located at its 48 cm mark, l = 48 cm

We need to find the magnitude of the torque if it is supported at the 25-cm mark, l' = 25 cm

It is given by :

`\tau=mgl`

`\tau=0.3\ kg* 9.79\ m/s^2* 0.23\ m`

`\tau=0.675\ Nm`

So, the magnitude of the torque is 0.675 N-m. Hence, this is the required solution.