Question

A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per second is 5 s is close to (A) 2.0 × 10⁻⁵ Nm (B) 4.0 × 10⁻⁶ Nm
(C) 1.6 × 10⁻⁵ Nm (D) 7.9 × 10⁻⁶ Nm

Answer 1

Answer:

refer to the attachment for the solution

Answer 2

The constant torque, that will make the system rotate about AB at 25 rotations per second is 5 s is close to:

(A) 2.0 × 10⁻⁵ Nm

This can be calculated as follows:

• Angular acceleration (α) refers to the angular velocity (ω) divided by acceleration time (t) and can be represented as-

       α=Δω​/Δt

         =(25×2π) / 5​

         =10π rad/sec²

• Now, torque about AB is -

        τ = (5​MR²/ 4) α

           =5​×5×10⁻³×(10⁻²)² x 10π

           =1.9625×10⁻⁵Nm

           ≃2.0×10⁻⁵Nm