Question

Two boxes, masses 10 kg and 8 kg, are attached to the end of a massless rod which pivots at 2 m from the left corner and held above the ground. The 10 kg mass is attached to the left end. The total length of the rod is 10 m.

1. Find the moment of inertia of the system around the pivoted point.

2. Find the net torque of the system when the massless rod is horizontal.

3. Calculate the angular acceleration of this system when it is first released.

Answer 1

1.

AB = 2 m

BC = 10 - 2 = 8 m

moment of inertia of the system is given as

I = M (AB)² + m (BC)²

I = (10) (2)² + (8) (8)² = 552 kgm²

2.

Net Torque is given as

= mg (BC) - (Mg) (AB)

= (8 x 9.8) (8) - (10 x 9.8) (2)

= 431.2 Nm

3.

angular acceleration is given as

= /I

= 431.2 /552

= 0.78 rad/s²