**Find the magnitude of the angular acceleration of system:**

A homogeneous sphere having a mass of 100 kg is associated to a slender rod having a mass of 20 kg. In the horizontal position, the angular speed of the system is 8 rad/sec. calculate the magnitude of the angular acceleration of the system and the reaction at O on the rod.

**Solution**

The system shall be rotating about O as shown in the figure. First let us try to locate the mass centre of the system.

∴ x¯ = (20 × 300 + 100 × 750) /120

= 675 mm .

The position of the mass centre is shown in the free body diagram in Figure (b). All the forces acting on the system are illustrated on FBD.

External forces are reactions at O, i.e. OT and ON and gravitational force mg shall be acting through the mass centre. r ω^{2} and r α shall be normal and tangential components of acceleration and mr ω^{2} and mr α are initial forces in these respective directions.

The total moment of inertia I equals the moment of inertia of the rod about its end ( 1/3) ml^{2} plus the transferred moment of inertia of the sphere, that means

( 2/5) mr ^{2 } + md ^{2}

where, d = 750 mm = 0.75 m

∴ I = (1/3) 20 (0.6)^{2} + (2/5) 100 (0.15)^{2 } + 100 (0.75)^{2}

= 59.55 kg m ^{2} .

Now apply dynamic equilibrium equations. ∑ M _{o} = 0 shall give us magnitude of α.

∴ ∑ M - I α = 0

∴ 120 × 9.8 × 0.675 = 59.55 α

∴ α = 13.3 rad / sec ^{2} .

Now, ∑ F_{x} = 0 . Thus, the sum of the forces on horizontal axis,

- ON + m r ω^{2 }= 0

∴ ON = 120 × (0.675) (8)^{2} = 5180 N to the left.

And,

∑ F_{y} = 0

- OT - mg + mr α = 0

∴ OT= mg - mr α

= 120 × 9.8 - 120 × 0.675 × 13.3

∴ OT = - 98.7 N.

Thus, direction of OT is upwards.